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Jflassarfjiusetts Institute of ^erijuologg 



NOTES 



ON 



GEAPHIC STATICS 



WITH APPLICATIONS TO 



TRUSSES, BEAMS, AND ARCHES 



BY 



JEROME SONDERICKER 



WOV 16.1896 



Notfcjooti Press 

J. S. CUSHTNG & CO., PRINTERS 

1896 



CONTENTS. 

CHAPTER I. 
General Methods. 

PAGE 

§ 1. Introduction 1 

§ 2. Funicular Polygon . . . . . . . . 5 

§ 3. Forces in Equilibrium ........ 8 

§ 4. Graphical Determination of Moments 11 

§ 5. Geometrical Properties of Funicular Polygons .... 13 

CHAPTER II. 

Roof Trusses. 

§ 1. Determination of Reactions of Supports 18 

§ 2. Determination of Stresses ........ 23 

§ 3. Counterbracing 35 

§ 4. Trusses with a Double System of Web Members . . .41 

§ 5. Three-hinged Arch ......... 42 

§ 6. Trusses loaded at Other Points than the Joints .... 49 

§ 7. General Directions . . . . . 51 

CHAPTER III. 

Beams. 

§ 1. Shearing Force and Bending Moment 54 

§ 2. Deflection of Beams 57 

§ 3. Beams supported at More than Two Points .... 60 

CHAPTER IV. 

Masonry Arches, Abutments, etc. 

§ 1. General Conditions of Stability ....:.. 64 

§ 2. Masonry Arch. Line of Pressure ..... 66 

§ 3. Abutments, Piers, etc. .... ... 74 



Jflag;saci)itj3ietts Institute of Eecl)noiogg 



NOTES 



ON 



GEAPHIC STATICS 



WITH APPLICATIONS TO 



TRUSSES, BEAMS, AND ARCHES 



JEROME SONDERICKER 



<<, 



vl 




NOV 16 1896 



RTntfoHroD Press 

J. S. CUSHING & CO., PRINTERS 

1896 



\ 







11° 



Copyright, 1896, 
By JEEOME SONDEEICKEE. 



J. S. Cushing & Co. — Berwick & Smith 
Norwood Mass. U.S.A. 



A. 






GRAPHIC STATICS. 

CHAPTER I. 

GENERAL METHODS. 
§ 1. Introduction. 

1. Graphic Statics has for its object the solution of problems in 
statics by means of geometrical constructions, the results being 
obtained directly from the scale drawings. 

The reader is assumed to be familiar with the principles and 
methods of statics as commonly presented in text-books on Mechanics. 
These will, however, be briefly stated here, for the case of forces 
lying in the same plane. 

2. Representation of Forces. A force is fully determined when 
its magnitude, direction and point of application are known. In deal- 
ing with problems in Statics of Bigid Bodies, the magnitude, 
direction and line of action of a force are the elements commonly 
involved, since the equilibrium or motion of such a body is not 
affected by transferring the point of application of a force to any 
other point of its line of action. 

3. Resultant of Any System of Forces lying in the Same Plane. 

The magnitude and direction of the resultant of any system of 
forces lying in the same plane may be found, I, by representing the 
given forces by the sides of a polygon taken in order, when the 
closing side in reverse order is the resultant, or, I a , by resolving 
each force into components parallel to coordinate axes and com- 
bining the resultants of the two sets of components thus formed, the 
components being treated as if all were applied at the same point. 

The line of action of the resultant may be found, II, by using the 
principle that the resultant of two forces lying in the same plane 
must pass through their point of intersection, or, II a , by the method 
of moments ; the moment of the resultant of any system of forces 
lying in the same plane being equal to the algebraic sum of the 
moments of the forces. 

1 



2 



NOTES ON GRAPHIC STATICS. 



In case the magnitude of the resultant is zero, the forces will 
either form a couple or be in equilibrium. If the resultant is a 
couple, its moment can be found by II a , or the given forces can be 
combined into a single resultant couple by II. 

If the lines of action of the given forces intersect at a common 
point, the line of action of the resultant will pass through this 
point, its magnitude and direction being found by the methods 
already stated. 

4. Examples. 1. Find the resultant of the four forces F, F', F", 
F'" (Fig. 1 A) by each of the preceding methods. 

First Solution. Represent the given forces by the sides of the 
polygon (Fig. 1 B) taken in order ; then the closing side, AD, repre- 
sents the magnitude and 
direction of the result- 
ant. The numerical val- 
ues of these quantities 
may be found, (1) by 
solving the polygon alge- 
braically, or (2) by di- 
rect measurement from 
a scale drawing. 

To find the line of 
action of the resultant 
by the first method, we can proceed as follows : The magnitude and 
direction of the resultant of F smd F' is AB. Its line of action, R', 
is drawn parallel to AB through the intersection of the lines of 
action of F and F'. Continuing in the same manner, we determine 
B" to be the line of action of the resultant of F, F', and F", and 
finally R to be the line of action of the resultant of the four 
forces. 

Second Solution. Eesolve each of the four forces into horizontal 
(H) and vertical (V) components. Then 




F=i5o 



Fig. 1. 



%H= 150 + 100 cos 45° + - 150 cos 30° = 90.8, 
% V = + 100 sin 45° + 80 + 150 sin 30° = 225.7, 



B = V(2#) 2 +(2F) 2 = 243.3, 
= 68° 5'. 



OL. = COS X/ » - 



R J 



INTRODUCTION. 3 

To find the line of action of R, apply the method of moments. 
Using as moment axis, we have 

tM = 150 • - 100 • 1.414 + 80 • - 150 • .866 = - 271.3. 

Hence, the moment of the resultant about O is left-handed, and its 
distance from O is — — == 1.115. This locates the resultant as given 

in Fig. 1. 

The moment arms of the several forces may be computed or may 
be measured directly from a scale drawing when the results thus 
obtained are sufficiently exact. 

Example 2. Assume four forces, not intersecting at the same 
point, which form a polygon. Find the resultant couple by each of 
the two methods mentioned in Art. 3. 

5. Conditions of Equilibrium : Forces not acting at the Same 
Point. The geometrical conditions of equilibrium for any system of 
forces lying in the same plane are, I, that the forces can be repre- 
sented in magnitude and direction by the sides of a polygon taken in 
order, and, II, that the line of action of the resultant of any portion 
of the given forces must coincide with the line of action of the 
resultant of the remainder. 

The algebraic conditions of equilibrium are, I a , that if the forces 
be resolved into components parallel to coordinate axes, the algebraic 
sum of each set of components must be zero, and, II a , that the alge- 
braic sum of the moments of the forces must equal zero about any 
moment axis perpendicular to the plane of the forces. 

Conditions I and I a are equivalent to each other ; also II and II a 
are equivalent. If I and I a , only, are satisfied, the resultant is a 
couple. 

Forces acting at the Same Point. In this case, condition I or I a is 
sufficient. The condition that if three non-parallel forces balance, 
they must intersect at a common point, is a special case under II. 

When a system of forces lying in the same plane is in equilib- 
rium, one or more of the preceding conditions of equilibrium serve 
to determine the unknown elements of the problem, if it is solvable 
under the assumption that the body acted on is rigid. 

6. Example. The portion of the truss (Fig. 2) to the left of AB 
is in equilibrium under the action of the supporting force P, load 
W, and the forces exerted by the portion of the truss to the right of 
the section AB upon the left-hand portion. The lines of action of 



NOTES ON GRAPHIC STATICS. 



these latter forces coincide with, the centre lines of the members cut 
by AB, and their magnitudes are equal to the stresses existing in 
these members. P and W being known, indicate how to find the 
three unknown forces by each of the following methods (see Art. 5) : 




(1) By using conditions II and I ; (2) by using II a alone ; (3) by 
using II a to find one force, then I to find the remaining two ; (4) by 
using II a to find one force, then I a to find the remaining two forces. 

First Solution. The resultant of P and F" acting at 0' must 
balance the resultant of W, F, and F' acting at 0. Hence the line 
of action of each resultant, R, must be 00'. P, F", and their 
resultant R, must form a triangle (Fig. 2 B) ; P being the known 
force, F" and R are thus determined. The resultant, R, must bal- 
ance the forces at 0, hence R, W, F, and F' must form a polygon as 
shown in Fig. 2 B. F and F', the remaining unknown forces, are 
thus determined. 

Second Solution. To find F, take the moment axis at the inter- 
section 0" of the other two unknown forces, so that their moments 
will each be zero. Then the algebraic sum of the moments of P, W, 
and F about 0" must equal zero. Solving the equation thus formed, 
we determine F. Similarly, to find F, take moments about 0', and 
to find F" take moments about 0. 

Third Solution. We find one force, as F, by the preceding 
method, then represent the known forces P, W, F, by the sides of a 
polygon taken in order, and complete the polygon by lines parallel 
to the two remaining forces F' and F". 

Fourth Solution. We find one force F by the method of moments 
as before. Then, placing the algebraic sums of the horizontal and 
vertical components of the forces each equal to zero, we form two 
equations which are solved for the two remaining unknown forces 
F' and F". 

In algebraic solutions the directions of the unknown forces are 
assumed, the algebraic signs of the results indicating whether the 
assumed directions are correct or not. 



FUNICULAR POLYGON. 




Fig. 



§ 2. Funicular Polygon. 

7. Definitions. Let F, F', F" (Fig. 3), be given forces, their 
magnitudes being represented by AB, BC, CD (Fig. 3 B). Assume 
any point P, and draw the 
radial lines PA, PB, etc. 
From any point M on the 
line of action of F draw 
ML and MN parallel to PA 
and PB respectively. From 
N, where MN intersects the 
line of action of F', draw 
NO parallel to PC; simi- 
larly draw OQ parallel to 
PD, thus forming the broken 
line LMNOQ. We may con- 
sider AP and PB, having 
the directions of the arrows 
marked 1, to be the com- 
ponents of the force F, the 
lines of action of these com- 
ponents being ML and MN respectively. Similarly, BP and PC, 
having the directions marked 2 and the lines of action NM and NO, 
may be taken as components of F'; and CP and PD with ON and 
OQ for lines of action as the components of F". MN is thus the 
line of action of two equal and opposite forces PB, and NO of the 
two equal and opposite forces PC. These two pairs of forces con- 
sequently balance, leaving AP and PD, having ML and OQ for 
their lines of action, as the equivalent of the original forces. 

The broken line LMNOQ is called a funicular or equilibrium 
polygon. The former name is given because the line corresponds to 
the shape assumed by a weightless cord when fastened at the ends 
and acted on by the given forces. This is shown by the polygon 
drawn in dotted lines. The latter name is applicable since a jointed 
frame of the form of the polygon would be in equilibrium under the 
action of the given forces. 

The point Pis called the pole; the lines PA, PB, etc., are the rays; 
and the corresponding lines of the funicular polygon are its strings. 

Figure 3 A is called the space diagram, since it represents the 
location of the lines of action of the forces. Its scale is one of dis- 
tance. Figure 3 B is the force diagram, the lengths of its lines rep- 
resenting the magnitudes of the forces to scale. The perpendicular 



6 NOTES ON GRAPHIC STATICS. 

distance from the pole to any side of the force polygon is called the 
pole distance of that force. It is to be noted that this distance 
represents a force magnitude. 

In the case of parallel forces the force polygon becomes a straight 
line, and the pole distances of all the forces are equal. 

8. Applications. The following results are readily derived from 
the construction explained in the preceding paragraph. 

(1) The resultant of F, F', and F" is given in magnitude and di- 
rection by the closing side AD of the force polygon, and its line of 
action passes through the point of intersection of the strings LM 
and OQ. The resultant is thus completely determined ; and, in 
general, the line of action of the resultant of any system of forces, taken 
consecutively, passes through the point of intersection of the two strings 
between which the forces lie. 

(2) If the force polygon is closed, PA and PD will coincide, and 
hence the corresponding strings LM and OQ will be parallel. In 
this case the resultant is a couple whose arm is the perpendicular 
distance between the parallel strings, and whose forces are repre- 
sented by the ray {PA = PD) corresponding to these strings. 

(3) In order for the given forces to be in equilibrium, the arm of 
the couple in (2) must be zero; that is, the strings LM and OQ must 
coincide in MO. In this case, the funicular polygon is said to be 
closed. The conditions of equilibrium, therefore, are that both the 
force and funicular polygons must close. 

If the forces intersect at a common point, the first of these condi- 
tions is sufficient, since such a set of forces cannot form a couple. 

(4) Any number of funicular polygons may be drawn for the 
same system of forces by using different poles and beginning the con- 
struction of the polygons at different points. Various geometrical 
relations exist between these different polygons, some of which are 
given in § 5. The following relations are derived directly from the 
preceding discussion. 

(a) Corresponding pairs of non-parallel strings of the various 
funicular polygons must intersect on the same straight line, this 
being the line of action of the resultant of the forces included by 
these strings. 

(b) If the force polygon is closed, and one funicular polygon closes, 
all funicular polygons for the given system of forces must close. 

In selecting the pole, the obtaining of accurate and convenient 
diagrams is kept in view. Generally the rays should not make very 
oblique angles with the adjacent lines of the force polygon. If 



FUNICULAR POLYGON. 



the pole is taken at a vertex of the force polygon, as at A, Fig. 3 B, 
the construction of the funicular polygon becomes identical with that 
explained in Art. 4, for finding the line of action of the resultant 
by the first method. The construction in Art. 4 frequently leads 
to inaccurate and inconvenient diagrams and is inapplicable to 
parallel forces. 

9. Notation. To illustrate the notation to be used, let it be 
required to find the resultant of the four forces ab, be, cd, de (Fig. 4). 
The line of action of any force 
is represented by the two letters *---.. 

between which it lies ; thus, ab 
represents the line of action of 
the first force, be of the second 
force, etc. In the force diagram, 
the same letters in capital type 
are used, but placed at the ends 
of the lines representing the forces. 
In constructing the force polygon, 
the forces and letters must be 
taken in order from the space dia- 
gram (e.g. proceeding from left to 
right). When this is done, if the 
letters are read in order from left 
to right in the space diagram, the same order of letters in the force 
polygon indicates the directions in which the successive forces act. 
It will be noticed that the string correspond- 
ing to the ray PA lies in the space a, that 
corresponding to PB in the space b, etc. These 
strings will be referred to as the strings a, b, 
etc. 

Having constructed the force and funicular 
polygons, the magnitude and direction of the 
resultant is represented by the closing side, 
AE, of the force polygon, and one point in 
its line of action is the point of intersection 
of the similarly lettered strings (a and e). The 
line of action of the resultant is then drawn through this point 
of intersection, parallel to AE. In referring to any force (e. g. the 
force BO), the capital letters will be employed, although the line 
of action of the force as well as its magnitude and direction is 
included in the reference. 




Fig. 4. 




Fig. 5. 



8 NOTES ON GRAPHIC STATICS. 

The system of notation just described is applicable in most cases 
occurring in engineering practice. When the location of the forces 
is such that it cannot be used to advantage, the modified notation 
illustrated in Fig. 5 may be employed instead. This needs no 
description. 

In general, in referring to forces in the text, the letters will be 
used so as to indicate by their order the direction in which the 
forces act. 

10. Problems. 

(1) Assume five non-parallel forces, and find their resultant, 
using two different poles. 

(2) Assume five non-parallel forces whose force polygon closes, 
and find their resultant, using two different poles. 

(3) Assume five parallel forces and find their resultant, using 
two different poles. 

(4) Assume five parallel forces whose algebraic sum is zero and 
find their resultant, using two different poles. 

(5) Assume five parallel forces which balance, and find their 
resultant, using two different poles. 

§ 3. Forces in Equilibrium. 

11. Use of Funicular Polygon. It will be noticed from § 2 that 
the funicular polygon construction is especially adapted to the case 
of parallel and other non-concurrent forces, although it may be used 
to advantage in problems relating to forces intersecting at a com- 
mon point, when this point lies outside the limits of the drawing. 
The conditions of equilibrium, given in Art. 8, for non-concurrent 
forces are (1) that the force polygon must close, and (2) that the 
funicular polygon must close. The application of these conditions to 
the solution of problems will now be illustrated, the cases most 
frequently arising in practice being presented. 

12. Case I. Parallel Forces. Given a system of parallel forces 
in equilibrium, the lines of action of all and the magnitudes and 
directions of all but two being known. It is required to find the 
unknown elements. 

Let the unknown forces be the supporting forces of the beam 
(Fig. 6). Represent the loads in succession by the sides AB, BC, 
CD of the force polygon, and, selecting a suitable pole, draw the 
strings a, b, c, d of the funicular polygon. The string a intersects 
the left reaction at m, and the string d the right reaction at n. 



FORCES IN EQUILIBRIUM. 



9 




a ^^ 

s^' i' ... 


c 


d 


' ^ 


Y \ 






e 




rii 



Fig. 6. 



The line joining m, n is then the closing side e of the funicular 
polygon, and the ray PE drawn parallel to it determines DE and 
EA to be the right and left 
hand supporting forces re- 
spectively. When the un- 
known (supporting) forces 
are parallel, but the given 
forces (loads) are not, the 
supporting forces must be 
parallel to the resultant load. 
If we take AD (Fig. 6) to represent this resultant load, the con- 
struction for determining the magnitudes of the supporting forces 
will be similar to that just explained. 

The method of determining the magnitudes and directions of 
the supporting forces from the lettering, after the construction is 
completed, should be carefully noted. It is as follows : 

(1) The strings are lettered with the same letters as the cor- 
responding rays, so that the strings intersecting on the line of 
action of any force have the same letters as those which represent 
that force in the force polygon. The two strings intersecting on the 
left supporting force are a and e, hence AE represents the magni- 
tude of this force. Similarly d and e intersect on the right support- 
ing force ; so its magnitude is DE. 

(2) The forces are laid off in the force polygon in right-handed 
order ; i.e. if the letters representing them in the space diagram be 
read in right-handed order, the same order of letters in the force 
polygon indicates the direction in which the forces act. We also 
know that the force polygon must close. The force polygon then is 
ABCDEA, the order of letters DE, EA indicating that each sup- 
porting force acts upwards. It should be noted in this and the 
following solutions that the known forces are made to follow consecu- 
tively in constructing the force polygon. 

(To obtain a thorough understanding of these solutions, it would 
be well for the student to trace out the construction from the stand- 
point of the triangle of forces. Thus, in the present case, the force 
AB is resolved into AP and PB, whose lines of action are the strings 
a and b. Similarly BC is resolved into BP and PC, and CD into 
CP and PD. The two forces PB and BP balance since they are 
equal and opposite and have the same line of action, b. PC and CP 
balance for a similar reason ; hence the three loads are equivalent 
to AP and PD, whose lines of action are a and d respectively. Now, 
in order to have equilibrium, the resultant of AP and the left sup- 



10 



NOTES ON GRAPHIC STATICS. 



porting force must balance the resultant of PD and the right 
supporting force. The lines of action of these two resultants must 
then coincide in mix. Since, of the three forces intersecting at m, 
one is the resultant of the other two, they must form a triangle, AP 
being the known magnitude. This triangle is APE ; thus the reac- 
tion EA is determined. The corresponding triangle for the forces 
intersecting in n is EPD.) 



13. Case II. Non-parallel Forces. Given a system of forces in 
equilibrium, all but two of which are known completely. Of these 
two, the line of action of one and one point in the line of action of the 
other are known. It is required to determine the unknown elements. 
Let the known forces be the wind pressures AB, BC, CD on the 
roof (Fig. 7). The unknown forces are the reactions of the sup- 
ports. The line of action of 
the right supporting force is 
given, it being vertical ; and 
one point, m, of the left reac- 
tion is known. Represent 
the loads in succession by 
the sides AB, BC, CD of the 
force polygon. The strings e 
and a must intersect on the 
line of action of the support- 
ing force EA, and as the 
point m of this line of action 
is known, the funicular poly- 
gon will be constructed so 
that these strings will intersect at this point. The construction 
of the funicular polygon is then begun by drawing the string a 
through the point m, the strings b, c, d being then drawn in 
order. 

The string d intersects the right supporting force at n, and, of 
course, the string a intersects the left supporting force at m ; mn is 
thus the closing side, e, of the funicular polygon. The ray cor- 
responding to the string e is now drawn, intersecting the right sup- 
porting force, DE, at E. The triangle of forces, PDE, whose lines 
of action intersect at n, is thus formed, and the magnitude of the 
supporting force DE is determined. Finally, the closing side, EA, 
of the force polygon represents the other supporting force in magni- 
tude and direction. Its line of action is drawn through m, parallel 
to EA. 




Fte. 7. 



DETERMINATION OF MOMENTS. 



11 




14. Case III. Non-parallel Forces. Given a system of forces 
in equilibrium, the lines of action of all are known, but the magni- 
tudes and directions of three of the forces are unknown. It is 
required to determine the unknown elements. 

Let ab, be, cd (Fig. 8) be the lines of action of 
the three unknown forces. The resultant of any 
two, as be and cd, must pass through their point 
of intersection, n. If this resultant be substituted 
for be and cd, the problem becomes identical with 
Case II ; ab being the line of action of one unknown force, and n 
one point of the line of action of the other. 

The construction of Case II having been made, the resultant of 
be and cd can be resolved into components parallel to these lines of 
action, thus completing the solution. Figure 9 shows this method 
applied to determine the stresses in the members ef, fg, and ga, of 
the truss, the loads and reactions being known. 



Fig. 8. 




Fig. 9. 



15. Resolution of Forces into Components. Since forces equal 
and opposite to the components and having the same lines of action 
will balance the given forces, we can solve problems of this nature in 
the same manner as if the forces balanced, the desired components 
being the balancing forces with their directions reversed. 



§ 4. Graphical Determination of Moments. 



16. Moment of a Given Force. Let AB (Fig. 10) be any force, 
and M any moment axis, distant x from the line of action of the 
force. Assuming any pole, P, resolve AB into the components AP 
and PB. Draw through M a line parallel to AB, and let PR be 



12 



NOTES ON GRAPHIC STATICS. 



two triangles 



PAB and OST being 




perpendicular to AB. The 

- similar, we have 

AB : PB : : ST : x, ov AB - x = PB - ST. 

But AB • x is the moment of the force about M, 
hence : 

The moment of a force about a given axis may be 
found by drawing a line through the moment axis par- 
allel to the given force, and multiplying the pole dis- 
tance of the force by the distance intercepted on this 
line by the two strings corresponding to the force. 

17. Moment of Resultant of Any System of Forces. 

The resultant of AB, BC, and CD (Fig. 11) is AD, its line of 
action passing through the intersection of the strings a and d. The 
moment of this resultant 
about any point M is there- 
fore equal to the intercept, 
ST, times the pole distance, P^rrr^TjR 
PR, of the resultant. It 
should be noted that in order 
to employ this construction 
the given forces must follow 
consecutively in the force 
polygon. The strings intercepting ST are lettered with the same 
letters as the resultant force. 




Fig. 11. 



18. Moment of Resultant of Parallel Forces. The constructions 
of this section are especially adapted to the case of parallel forces. 

Illustration: Let the beam (Fig. 12) be 
loaded as shown. The reactions of the 
supports (Art. 12) are found to be CD and 
DA. It is required to find the bending 
moment at any section, ST. By definition, 
this bending moment is equal to the alge- 
braic sum of the moments of the forces to 
the left of the section, and, by Art. 17, this 
resultant moment is equal to the pole dis- 
tance, y, times the intercept ST. 

Now in case of parallel forces, the pole 
distance is the same for all the forces, 
hence the bending moments at the various sections of the beam are 




GEOMETRICAL PROPERTIES OF FUNICULAR POLYGONS. 13 

directly proportional to the corresponding intercepts. This sub- 
ject is further developed in Chapter III. 

§ 5. Geometrical Properties of Funicular Polygons. 

19. Funicular Polygon having the Direction of One String given. 

Let it be required to draw a second funicular polygon (Fig. 12) 
such that the string d will be horizontal. Since this string must be 
parallel to the ray PD, if we draw a horizontal line through D, a 
polygon constructed from a pole taken at any point on this line will 
fulfil the requirement. 

20. Intercepts inversely Proportional to Pole Distances. Parallel 
Forces. It follows directly from Arts. 17 and 18 that if any two 
funicular polygons be constructed for a given system of parallel 
forces, the ratio of the intercepts of corresponding pairs of strings 
on any line drawn parallel to the forces is constant, being equal to 
the inverse ratio of the pole distances of the respective polygons. 

21. Locus of Points of Intersection of Corresponding Strings. 

If two funicular polygons be constructed for the same system of 
forces, their corresponding strings will intersect on a straight line 
parallel to the line joining the two poles. 

Proof. Let AB, BC, CD (Fig. 13) be the given forces, P and 
P' being the two poles. The four forces AP, PB, BP', PA balance, 
since the resultant of AP and PB, i.e. AB, is equal and opposite and 
has the same line of action 
as the resultant of BP' 
and PA. Since these 
forces balance, the result- 
ant of either two, as AP 
and PA, must balance the 
resultant of the other two, 
PB and BP'. The result- 
ant of AP and PA is 

P'P, and its line of action 

i FlG - 13 - 

must pass through the 

intersection x of the corresponding strings. Similarly, the resultant 
of PB and BP f is PP', and its line of action must pass through y. 
Since these two resultants balance, their lines of action must coin- 
cide in xy, which is therefore parallel to PP. The intersection of 
the remaining pairs of strings on xy can be proved in a similar 
manner. 




14 



NOTES ON GRAPHIC STATICS. 



22. Locus of Poles of Funicular Polygons passing through Two 
Given Points. It follows directly from the preceding proof that if 
any other funicular polygon be drawn for the given system of forces 
such that the string a would pass through x and the string b through 
y, the corresponding pole would lie on the line PP', and in general : 
For any given system of forces, the locus of the poles of all funicular 
polygons, two of whose corresponding strings pass through two given 
points, is a straight line parallel to the line joining the two points. 

23. Funicular Polygon through Three Points. General Solution. 

Problem. To draw a funicular polygon for a given system of 
forces such that three designated strings shall pass through three 
given points. Let the forces be AB, BC, CD, DE, EF (Fig. 14). It 
is required to draw a funicular polygon such that the string a will pass 
through 0, the string c through 0', and the string/ through 0". Let 




Fig. 14. 



mn be any funicular polygon for the given forces, with P for pole. 
Through O and 0' draw lines parallel to AC. These may be taken as 
the lines of action of two forces which balance AB and BC. To 
determine their magnitudes draw the closing string px and its 
corresponding ray PX, giving CX and XA for the required balancing 
forces. (See Art. 12.) Now, in order for the strings a and c to pass 
through and O' respectively, the closing stringy must pass through 
both these points, hence the corresponding pole must lie on the ray 
P'X, drawn through X parallel to 00'. (See Art. 19.) If any point 
on this line be taken for pole and a funicular polygon drawn such 
that the string a passes through 0, the string c will pass through O'. 
In a similar manner P'X' is determined as the locus of the poles of all 
funicular polygons whose strings c and / pass through 0' and 0" 
respectively. The point of intersection P! is then the pole of the 
required polygon, which is drawn. Evidently a similar construction 



GEOMETRICAL PROPERTIES OF FUNICULAR POLYGONS. 15 

could be made for the points and 0". Another method of solving 
this problem is given in Art. 52. 

24. Funicular Polygon through Three Points. Parallel Forces. 

A shorter solution than that given in Art. 23 can be made for this 




Fig. 15. 



case. Let AB, BC, CD, DE (Fig. 15) be the given forces, and O, 0', 
and 0" the given points. Let mn be any funicular polygon for the 
given forces, P being its pole. Draw lines parallel to the forces 
through each of the three given points. Assume two of these lines, 
as the ones through and 0", to be the lines of action of forces 
balancing the given forces included between them. The magnitudes 
of these balancing forces are found by drawing the closing string 
and parallel to it the ray PF, as in Art. 23. Now, in order for the 
strings a and e to pass through and 0" respectively, the closing 
string must be the line 00", and the corresponding ray FP' will con- 
tain the new pole. The distance intercepted on the line drawn 
through 0' by the first polygon is PS, and by the second polygon 
will be O'S'. The new pole distance, y', can then be found from the 
proportion y' : y : : PS : O'S' (Art. 20). This locates the pole P' of 
the required polygon. 

Second Method. Instead of using the pole P' in constructing the 
second polygon, its vertices may be located directly from the fact 
that the ratio of corresponding intercepts of the two polygons is con- 
stant and hence equal to the ratio PS : O'S'. (See Art. 20.) 

25. Funicular Polygon through Two Points, One String having a 
Given Direction. Problem. To draw a funicular polygon for a given 
system of forces, such that two designated strings shall pass through 
two given points and one string of the polygon shall have a given 
direction. 

The method of Art. 23 can be used to determine the locus of the 
poles of all polygons passing through the two given points. The 



16 



NOTES ON GRAPHIC STATICS. 



intersection of the ray corresponding to the string whose direction is 
given with this locus will be the pole of the required polygon. 

If the string whose direction is given is also one of the two which 
are to pass through the given points, the following simpler solution 
can be made. Let AB, BC, and CD (Fig. 16) be the given forces, 

and P the pole of any funic- 
ular polygon mn for these 
forces. • It is required to 
draw a second polygon such 
that the strings a and d will 
pass through and 0' re- 
spectively, and the string a 
be horizontal. The resultant 
of the ^forces included be- 
tween £fre designated strings 
a and d is AD, its line of 
action passing through the intersection S of the strings a and d of 
the first polygon. The corresponding strings of the second poly- 
gon must intersect on this line (Art. 8). These strings are, there- 
fore, OS', drawn horizontally, and S'O'. The new pole P' will be at 
the intersection of the corresponding rays. The polygon can now 
be completed. 

26. Remarks. In constructing a funicular polygon for a given 
system of forces, the strings are drawn in succession, the correctness 
of the location of each depending upon the accuracy with which all 
the preceding ones have been drawn. In this way errors accumulate, 

i5 




Fig. 16. 




Fig. 17. 



and it is necessary to make the construction with extreme care. 
When, however, one polygon has been drawn, any string of a second 
polygon can be located directly. 



GEOMETRICAL PROPERTIES OF FUNICULAR POLYGONS. 17 

Three special methods of drawing any string of a second polygon 
are shown in Fig. 17. The pole of the given funicular polygon is P. 
It is desired to draw any string cl of a second polygon, the pole 
being P' and the first string having the position p 'a. 

First Method. The two strings a intersect at 0. Draw OL 
parallel to PP'. This line will contain the points of intersection of 
the corresponding strings of the two polygons (Art. 21). The string 
d' is then drawn through the point of intersection of the string d 
with OL, and parallel to the ray P'D. 

Second Method. The strings a and d intersect at R. A line 
through this point parallel to AD is the line of action of the result- 
ant of the three forces included between these strings. The cor- 
responding strings of the second polygon must intersect on this 
line. The string a' is known, so d' can at once be drawn through 
the point of intersection, x\ of a' and the line through R. 

Third Method. Draw any line ST parallel to the resultant force 
AD. The strings a and d intercept the distance ST on this line. 
The moment of the included forces AB, BC, and CD, about any 
point on this line as moment axis, is then ST • y (Art. 17). This 
product must evidently be the same for both funicular polygons. 
The string a' intersects ST at S'. One point of the string d' is 
then located by laying off S'T', such that S'T' -y' = ST-y. A line 
through T' parallel to the ray P'D is the required string. 

The third method is especially adapted to the case of parallel 
forces. (See Art. 24, Second Method.) 



CHAPTER II. 

ROOF TRUSSES. 

(For construction of roofs, estimation of loads, etc., see Lanza's Applied 
Mechanics, Chap. III.) 

§ 1. Determination of Reactions of Supports. 

27. Direction of Supporting Forces. When a truss is fixed at 
both ends, the reactions of the supports are indeterminate. They 
are commonly assumed to be parallel to the resultant load. This 
assumption, however, is in some cases hardly admissible. (See 
Art. 28.) When one end is supported on rollers, thus providing for 
free expansion and contraction, the reaction . at the roller end is 
assumed to be vertical, the direction of the reaction at the fixed end 
being determined from the conditions of equilibrium. 

28. Examples. The methods of determining the supporting 
forces are included under Arts. 5, 12, and 13. A few examples will 
be given to indicate suitable solutions. 

Example 1. The unsymmetrical truss (Fig. 18) is subjected to 
a dead load uniformly distributed over the upper chord, and snow 
load on the right side. Find reactions of supports. 




Fig. 18. 



The loads supported at the joints are represented in order in the 
force polygon. The resultant loads on the two sides are AX and 

18 



ROOF TRUSSES. REACTIONS OF SUPPORTS. 



19 




XG, their lines of action being given by the dotted lines. The 
method of Art. 12 is used to determine the reactions, which are 
GH and HA. The two resultant loads are used instead of the 
loads at the joints, so as to reduce the number of sides of the 
funicular polygon, thus saving labor and increasing the accuracy 
of the result. 

Example 2. The truss (Fig. 19) is subjected to wind pressure 
from the left, the load being 
uniformly distributed over the 
rafter. Find reactions of sup- 
ports : (1) When both ends are 
fixed (Fig. 19 A) ; (2) when the a 
right end is supported by a 
roller (Fig. 19 B) ; (3) when 
the left end is supported by a 
roller (Fig. 19 C). 

The method of Art. 12 is 
used for the first case, and 
the reactions are also found 
by dividing the load AD 
into parts inversely propor- 
tional to the segments into 
which the line of action of the 
resultant load divides the line 
joining the supports. The 
method of Art. 13 is used for 
the second and third cases. 
In addition, the lines of action of the supporting forces are also de- 
termined from the condition that the resultant load and supporting 
forces (three forces in equilibrium) must intersect at a common 
point (Art. 5). 

Example 3. The truss (Fig. 20), fixed at the ends, is loaded 
with a vertical load at each joint of the upper chord, and also with 
a wind load on the left side. Find reactions of supports for this 
combined loading. The resultant wind pressure on segments b and 
c acts at their middle points, and this pressure in each case is sup- 
ported equally at the adjacent joints. At the joint be, for instance, 
three loads are shown : one-half the pressure on surface b, one-half 
the pressure on surface c, and the vertical load supported at the 
joint. 

The loads at the joints are represented in order in the force 
polygon. The lines of action of the supporting forces are drawn 




Fig. 19. 



20 



NOTES ON GRAPHIC STATICS. 



parallel to the closing side, AF, of the polygon. The lines of 
action of the resultant loads at the joints are also drawn parallel to 
the lines AB, BC, etc., of the force polygon. These are shown by 
dotted lines. A funicular polygon is now drawn for these resultant 
joint loads, and the magnitudes, FG and GA, of the supporting 
forces are determined as in the preceding cases. 

It is to be noted that if the supporting forces (Fig. 20) had been 
determined for the dead and wind loads separately, in each case 
assuming that the supporting forces are parallel, the resultant 




Fig. 20. 



reactions would not be parallel, and would therefore differ from 
those found in Fig. 20. This difference in results obtained by 
dealing with the separate and combined loads does not occur under 
any of the modes of treatment discussed other than that of parallel 
reactions. 

Example 4. Another assumption employed in determining the 
supporting forces for a truss with fixed ends is that the horizontal 
component of the loads is divided equally between the two supports. 
The solution of this case is illustrated in Fig. 21. AB is the result- 
ant load. The supporting forces are resolved into vertical and 
horizontal components, and the horizontal components, EA and BC, 
are assumed equal, and hence each is equal to one-half the horizontal 
component of the load. FA and BC can then be found by draw- 
ing a vertical, CE, through the middle point of AB. CE also 
represents the sum of the vertical components of the reactions. 
These components, CD and DE, can then be found by dividing CE 



ROOF TRUSSES. REACTIONS OF SUPPORTS. 



21 



into parts inversely proportional to the segments of the line joining 
the supports, or can be found by the funicular polygon. This latter 
construction is shown in the diagram. The resultant supporting 
forces are then BD and DA. 




Fig. 21. 

Of the two solutions for trusses with fixed ends, illustrated in 
examples 3 and 4, the latter would appear to be the more generally 
applicable. For comparatively flat roofs, either solution would seem 
to be admissible. 

Example 5. Figure 22 represents a truss supported on columns. 
The columns are assumed to be hinged at both ends, and the knee- 




Fig. 22. 



braces xy, x'y' prevent distortion under the action of non- vertical 
forces, such as wind pressure, the force exerted by belts, jib cranes, 
etc. The sides of the building are assumed to be covered with 
corrugated iron. This covering is supported by the columns, so that 
the frame is called upon to resist the pressure of the wind on the 



22 NOTES ON GRAPHIC STATICS. 

side as well as roof of the building. It is required to determine the 
external forces acting on truss and columns, due to wind pressure. 

Distance between trusses = 16 ft. 

Assumed wind pressure on vertical surface =30 lbs. per sq. ft. 

Normal wind pressure on roof (Hutton's formula) = 19 lbs. per sq. ft. 

Total pressure on side of building (one panel) = 16 x 40 x 30 = 19200 lbs. = P. 

Total normal pressure on roof (one panel) =16 x 34 x 19=10336 lbs.=P'. 

Horizontal component of P' =4864 lbs. 

Vertical component of P' =9120 lbs. 

The supporting forces are indeterminate. The assumption made 
by Hutchinson x will usually give a greater stress in the leeward 
than in the windward knee-brace. A more generally satisfactory 
treatment would appear to be to assume that the horizontal reactions 
are divided between the two columns in such a manner as to cause 
equal stresses in the two knee-braces. The example is solved on this 
latter assumption. 

Taking moments about top of left-hand column, we have 

Moment of stress in xy = H • 40 - 19200 ■ 20. 

For the right-hand column, 

Moment of stress in x'y' = H' • 40. 

Equating the second members of these two equations, we have 

H-H'^9600. 
From %H = we have 

H+H' = 19200 + 4864 = 24064. 

.-. H= 16832 lbs. and H' = 7232 lbs. 

The vertical reactions can now be found by moments. Taking 
the foot of the right-hand column as moment axis, we have 

V- 60 = 19200 • 20 + 4864 . 48 - 9120 . 45. 

.-. V= 3451 lbs. and V = 3451 + 9120 = 12571 lbs. 

The horizontal forces at y and z which balance H and P are found 
by moments to be ^=28928 lbs. and ^ = 31296 lbs. We can 
now determine the forces which act upon the column, and those 
which act on the truss, by applying two equal and opposite forces, 
P==28928 lbs. at y, and two equal and opposite forces, P 7 2 =31296 lbs. 
at z. Also apply at y two equal and opposite vertical forces, 
F x = V= 3451 lbs. These added forces will not affect the stresses. 

1 See Johnson's Modern Framed Structures, Chap. XXIX., or Proceedings 
Engineers' 1 Society of Western Pennsylvania, Vol. VIII., p. 247. 



ROOF TRUSSES. DETERMINATION OF STRESSES. 23 

We now have acting on the column a set of balanced forces, H, 
P, — F, and F 2 , causing a bending stress ; also the balanced forces 
V and F 1 causing tension in the lower portion of the column. 

The forces acting on the truss, taking y and y' as the points of 
support, are : at y, F and — F 1 ; and at z, — F 2 . The right-hand 
column is dealt with similarly, the dotted arrows giving the forces 
acting on the column, and those drawn in full lines, the forces act- 
ing on the truss. 

The polygon of external forces for the truss is plotted to scale 
in Fig. 22, HI= 21696 lbs., IJ= 28928 lbs. 

If the columns are rigidly fixed at the base, the fastening will 
assist the knee-braces in resisting distortion, and the stresses in 
knee-braces and truss will be reduced. 1 

When a truss is loaded at both upper and lower chords, as in 
Fig. 1, Plate I, a difficulty occurs in using the funicular polygon 
construction to determine the supporting forces. In representing 
the external forces in the force polygon, they are laid off in succes- 
sion, proceeding around the truss from left to right, i.e. in the order 
AB • • • UK • • • RA. This is done so as to be able to determine the 
stresses conveniently. On the other hand, in determining the sup- 
porting forces by the usual funicular polygon construction, the 
known forces must follow consecutively. This would require the 
construction of a second force polygon. No difficulty will be ex- 
perienced in finding the reactions by other methods in such cases. 

29. Algebraic Methods. Frequently the supporting forces can 
be computed readily, using the algebraic methods of Art. 5 or solv- 
ing the geometrical figures of the graphical constructions. Such 
computations should be employed freely in connection with graphi- 
cal solutions, for the purpose of securing a greater degree of accuracy, 
and as checks on the constructions. 

Problem 1. Indicate in detail a suitable method for computing 
the reactions of the supports (Fig. 18). 
Problem 2. Do the same for Fig. 20. 

§ 2. Determination of Stresses. 

30. General Methods. A truss is designed to support loads 
applied at the joints, by virtue of the resistance to extension and 
compression of its various members. 

There are two general methods for determining the tension and 

1 See Johnson's Modern Framed Structures, p. 159. 




24 NOTES ON GRAPHIC STATICS. 

compression stresses in the members of a truss : (1) method of sec- 
tions; (2) method of joints. 

31. Method of Sections. Let the imaginary line xy (Fig. 23) 
divide the truss into two parts, this line intersecting the three mem- 
bers CI, IH, HG. The portion of the 
truss to the left of xy is a body in equi- 
librium under the action of certain forces. 
Consider the member CI. If it is in 
tension, the portion to the right of xy 
must be exerting force upon the por- 
tion to the left of xy, toward the right. 

If CI is in compression, its right-hand portion must, on the other 
hand, be exerting force upon its left-hand portion, towards the 
left. In either case, the magnitude of the force is equal to that 
of the stress in the member, and the line of action of the force has 
the direction of the length of the member. This force is external 
with reference to the portion of the truss to the left of xy. It 
will be represented by the letters CI. Similar explanations hold 
for IH and HG. The forces, therefore, which hold the left portion 
of the truss in equilibrium are the known forces GA, AB, BC, and 
the forces, CI, IH, and HG exerted by the right-hand portion of 
these three members upon their left-hand portion. The lines of 
action of these last three forces are known, their magnitudes and 
directions being unknown. These six forces constitute a system of 
forces in equilibrium, lying in the same plane but not acting at the 
same point. One or more of the conditions of equilibrium of Art. 5 
can therefore be used to determine the unknown forces, as was indi- 
cated in Art. 6. Of these solutions, the method of moments is gen- 
erally most useful in dealing with roof trusses. 
The following points should be noted : 

(1) Only three forces, unknown in magnitude, can be determined, 
so that if the section xy cuts more than three members which are 
in action under the given loads, a solution cannot be made. 

(2) The stress in a member is equal to the magnitude of the force 
which it exerts, and the nature of the stress, tension, or compression 
can be determined from the direction of the force (tension, if directed 
away from left portion, otherwise compression). 

32. Method of Joints. The external forces acting upon the joint 
A (Fig. 23) are: (1) the supporting force GA-, (2) the load AB; (3) 
the forces exerted by the members BH and HG upon the joint. If 



ROOF TRUSSES. DETERMINATION OF STRESSES. 25 

either member, as BH, is in tension, the force which it exerts on the 
joint A is evidently directed away from the joint ; if in compression, 
towards it. As the joint is in equilibrium these four forces must 
balance, and by applying either the algebraic method of resolution 
of forces or the geometric method of polygon of forces, the unknown 
forces can be determined. 

In using the method of joints, the following points should be 
noted : 

(1) The forces dealt with are those acting on the joint. 

(2) In dealing with any one joint only two unknown forces can 
be determined. 

(3) The nature of the stress in a member can be determined from 
the direction of the force which the member exerts on the joint ; ten- 
sion, if the force acts away from the joint; compression, if the force 
acts towards the joint. 

(4) When the stress in a member is determined, the force which 
it exerts upon the joint at each end is known, these forces being 
equal and opposite. 

33. Determination of Stresses in Roof Trusses. Of the methods 
which have been explained, the one best adapted to roof trusses is 
the method of joints, solving by the polygon of forces. In applying 
this method, the external forces being known, we begin by construct- 
ing the polygon of forces at any joint of the truss where only two 
stresses are unknown. Having thus determined these two, we repeat 
the construction for another joint where only two stresses remain 
unknown, and continue in this manner until the stresses in all the 
members have been determined. 

34. Example. Bow's Notation. The truss (Fig. 24) is subjected 
to a uniformly distributed load, W. Each intermediate joint of the 
upper chord supports \W, and each end 
joint, \W. Each supporting force is \W. 
The external forces are lettered as in the 
preceding chapter, ab representing the 
left reaction; be, eel, etc., the successive 
loads. Letters are also placed in the 
spaces into which the surface is divided 
by the web numbers. Each member of 
the truss is represented by the letters in 
the adjacent spaces ; e.g. the two halves 
of the lower chord are ka and ha ; the 
vertical member is ji, etc. The direc- fig. u. 




26 NOTES ON GRAPHIC STATICS. 

tions of the forces are indicated in the force or stress diagram by 
the order of the letters. For this purpose the letters are to be read 
in right-handed order around the truss or any joint of it ; e.g. the left 
reaction is ab, left-hand load be, etc. ; the force exerted by the mem- 
ber ck upon the left-hand joint is ck, and upon the joint cdjk is ke. 

To illustrate the manner of using this notation, the stresses in 
the various members of the truss will be determined. First con- 
struct the polygon of external forces. Lay off the loads on a verti- 
cal line in right-handed order. The first load, beginning at the left, 
is be ; it acts downwards, and to indicate this, the force is lettered so 
that BC reads downwards. Continuing in this manner, when the* 
polygon is completed, if we read the letters surrounding the truss in 
right-handed order, bedefgab, these letters in the stress diagram, read 
in the same order, will represent the polygon of external forces, the 
order of succession of the letters indicating the direction in which 
the forces act. 

Next the polygon of forces for the left-hand joint of the truss is 
constructed. The known forces ab and be are already included in 
the stress diagram. The two unknown forces are ck and ka. One 
letter of each (C and A) is already in the stress diagram. These 
letters indicate the points from which to draw the lines representing 
the unknown forces. Thus the polygon is completed by drawing 
from G and A lines parallel to ck and ka respectively, their point of 
intersection being lettered K. Proceeding to the next joint, kc and 
cd are the known forces. From the points D and K, lines are drawn 
parallel to dj and jk respectively, intersecting at J f thus completing 
that polygon. By this mode of construction, the polygons of forces 
for the various joints of the truss are grouped together in a single 
diagram called the stress diagram. When the last joint is reached, 
all the stresses but one will be known. The letters representing this 
one will already be in the stress diagram, and the line joining them 
will be parallel to the corresponding member of the truss if the con- 
struction is accurate. 

The following points should be noted in using Bow's notation : 

(1) The stress in any member of the truss is represented by the 
same letters as the member itself. 

(2) The polygon of forces for any joint is lettered with the letters 
surrounding that joint, the direction of the forces being indicated 
by the succession of letters obtained by reading the letters surround- 
ing the joint in right-handed order. 

(3) To determine whether any member is in tension or compres- 
sion, read the letters representing that member in right-handed order 



ROOF TRUSSES. DETERMINATION OF STRESSES. 27 

about the joint at either end of the member. The same order of 
succession of these letters in the stress diagram indicates the direc- 
tion of the force which the member exerts on the joint used. If this 
direction is towards the joint, the stress is compression ; if away 
from it, tension. As an illustration, let the nature of the stress in 
ck be determined. Using the left joint, the letters read c-k. In 
the stress diagram, this order of succession of these letters is towards 
the left. Eeferring again to the truss, the direction, towards the 
left, is seen to be towards the joint used as a centre, this indicating 
that the member ck is exerting force on this joint towards it. Hence 
ck is in compression. 

(4) In constructing the polygons of forces it is to be noted that 
all the sides of any polygon but the two corresponding to the two 
unknown forces at the joint will be already represented in the stress 
diagram, and the points from which to draw these two sides are 
indicated by the lettering, as previously explained. The point of 
intersection of these, two sides is marked with the letter common 
to the corresponding members of the truss. 

In using the method illustrated, the truss is drawn accurately to 
scale, the loads are also represented accurately to scale, and the 
stress diagram is constructed with extreme care. The magnitudes 
of the stresses are gotten by scaling off the proper lines of the stress 
diagram, and the nature of the stress is derived from the lettering, 
as previously explained. The most serious cause of inaccuracy, 
especially in case of trusses having a large number of members, 
is that the construction of each polygon in turn is based upon pre- 
ceding ones, so that errors accumulate. To guard against this, it 
is well to determine the stress in one or more of the members 
analytically. The method of moments is especially adapted to 
such uses. (For detailed directions regarding the work of con- 
structing the diagrams, see § 7.) 



35. Example 1. Wooden Truss fixed at Both Ends. Figure 1, 
Plate I, represents one of a series of parallel trusses, spaced 16 
feet between centres, supporting a roof. The vertical iron tie-rods 
divide it into 8 panels of equal width. Span = 80 ft. Eise of 
upper chord = 16 ft. The roof is to be covered with tin laid on 
sheathing 1 in. thick. The sheathing is supported by rafters 2 in. 
by 7 in. section, spaced 2 ft. between centres. The rafters are 
supported by purlins 8 in. by 12 in. section, these being supported 
at the joints of the upper chord. 

Each rafter supports an area of the roof surface 2 ft. wide by 



28 NOTES ON GRAPHIC STATICS. 

10.8 ft. long, and is proportioned as a beam to support the maximum 
load, including its own weight, which is liable to come upon this 
area. Each purlin supports an area 10.8 ft. wide by 16 ft. long, 
and is also proportioned as a beam for the maximum load to which 
it is liable to be subjected. The student should verify these dimen- 
sions of rafters and purlins, using a working stress of 1000 lbs. per 
sq. inch. 

The snow load is assumed to be 20 lbs. per sq. ft. of horizontal 
projection of roof surface, and the wind pressure 40 lbs. per sq. ft. 
on a vertical surface. 

Calculation of Loads. 

Wt. of tin and sheathing per sq. ft. of roof surface = 3| lbs. 
" " rafters (30 lbs. per cu. ft.) " = \\ " 

" " purlins (30 lbs. per cu. ft.) " =2 " 

Total = 7 lbs. 

Estimated weight of truss, W= \ al (1 -f ^ I) = 5760 lbs. This 
weight is assumed to be supported at the joints of the upper chord. 

Normal component of wind pressure (Hutton's formula) =20 lbs. 
per sq. ft. of roof surface. 

(The weight of the truss must be estimated from the actual 
weights of trusses already built. Merriman gives the formula 
for wooden trusses, W = -J- al (1 -f- y 1 -^ l) 9 and for iron trusses, 
W= faZ (1 -f yVO- These formulas are derived from a table of 
weights of trusses given by Bicker. Johnson gives, for iron trusses, 
the formula, W= ■£% al 2 . In each case W= weight of truss in pounds, 
a = distance between trusses in feet, I = span in feet. These for- 
mulas can, at the best, be only rough approximations.) 

Loads at Intermediate Joints of Upper Chord. 

Roof covering, etc. 7 x 16 x 10.8 = 1210 lbs. 

Truss i x 5760 = 720 " 



Total dead load at each intermediate joint = 1930 lbs. 

Snow " " " " " = 20 x 16 x 10 = 3200 " 

Wind " " " ." " =20x16x10.8 = 3460 " 

Each extreme joint supports one-half the load carried by an inter- 
mediate joint. 

The roof is also to support a ceiling which is suspended from the 
joints of the lower chord. Its weight, estimated in the same manner 
as the preceding, is 17 lbs. per sq. ft., or 2720 lbs. per joint. 



ROOF TRUSSES. DETERMINATION OF STRESSES. 29 

36. Stresses. The general method of determining the stresses 
has already been explained. The student should trace out the 
construction of the diagrams. 

Dead Load Stress Diagram. The reaction of each support is 
\ (1930 x 8 + 2720 x 7) = 17240 lbs. The loads and reactions are 
laid off in right-handed order, and the polygons of forces for the 
various joints are constructed, beginning at the left support. Since 
the truss is symmetrical and symmetrically loaded, the stresses in 
corresponding members of the two halves will be equal, so the 
stresses for one side only are determined. The following special 
precautions were taken to ensure accuracy : 

(1) The intersection of BS and RS is quite oblique, and since the 
accuracy of the diagram depends upon the correctness of location 
of the point S, it is well to determine that point by computation 
as follows : The forces at the left support form a triangle similar 
to that formed by the half truss. Hence, 

RA - AB (16275 lbs.) : SB : : 16 : 40. .-. jSR = 40687 lbs. 

This serves to locate S accurately. The remaining stresses are then 
determined graphically. 

(2) In order to check the completed diagram, the stress in zf is 
computed by moments. (See Art. 6, Second Solution.) Taking the* 
middle lower joint as moment axis and dealing with the left-hand 
forces, we have 

FZ x 14.856 = 16275 x 40 - 4650 x 30 - 4650 x 20 - 4650 x 10. 

.:FZ = 25040 lbs. 

The value of this stress as measured from the diagram is 25000 lbs. 

Snow Stress Diagram. This diagram needs no explanation. The 
same precautions were taken as in the preceding case. The com- 
puted stress in zf is 17232 lbs., the diagram scaling 17200 lbs. 

Wind Sti*ess Diagram. This is drawn for wind on the left side. 
The total wind pressure is 4 x 3460 = 13840 lbs. As both ends 
are fixed, the reactions are assumed to be parallel to the resultant 
wind pressure, and their magnitudes, computed by the principles 
of parallel forces, are 9818 lbs. and 4022 lbs. The force polygons 
are constructed, working from each end towards the centre of the 
truss. It is to be noted that the stresses in all the web members 
on the lee side are zero. The diagram checks itself by the clos- 
ing of the last polygon of forces. All complete diagrams afford 



30 NOTES ON GRAPHIC STATICS. 

such a check. In addition, the stress in nl was found by moments 
to be 7827 lbs., the diagram giving 7800 lbs. 

37. Maximum Stresses. The members of a truss are proportioned 
to withstand safely the maximum stresses to which they are liable 
to be subjected. Having determined the stresses resulting from the 
dead load, wind, snow, etc., the different combinations of these loads 
liable to occur must be decided upon, and the corresponding stresses 
found by taking the algebraic sum of the stresses due to the sepa- 
rate loadings. The maximum result is then the stress to be used 
in estimating the dimensions of the members. The algebraic mini- 
mum stress should also be noted. If the stress in a member changes 
sign, the member must be designed to resist both tension and com- 
pression, also both extreme stresses are used when the working 
stress is determined in accordance with the laws of failure under 
repetition of stress. 

If it be assumed that maximum wind and snow loads cannot occur 
at the same time, we would have the following systems of loads to 
consider : (1) dead load ; (2) dead and snow loads ; (3) dead and 
wind loads. If the truss is unsymmetrical, or supported differently 
at the two ends, the dead load must be combined with wind load on 
each side. In the design of the arches supporting the train-shed 
•roof of the Philadelphia and Reading Terminal Railway, Philadel- 
phia, the stresses were determined for : (1) dead load ; (2) dead 
load and snow on one side ; (3) dead load and snow on both sides ; 
(4) dead load and wind on one side; (5) Dead and snow loads with 
wind on one side. 

38. Tabulation of Stresses. The stresses are scaled off from the 
diagrams and inserted in the table, page 31. The sign of the stress 
is determined from the lettering. (See Art. 34.) The combinations 
of stresses considered are : (1) dead load ; (2) dead and snow loads ; 
(3) dead load and wind on either side ; (4) dead and snow loads 
with wind on either side. The maximum and minimum results are 
recorded in the table. In this example, the maximum stresses for 
corresponding members on the two sides of the truss are evidently 
equal. 

For convenience of reference the stresses may also be given as 
shown in Fig. 1 d, Plate I. The double lines indicate compression 
numbers. 

39. Example 2. Fig. 2, Plate I, is an iron truss, one end resting 
on rollers. (In practice rollers would not be used for so short a span.) 



ROOF TRUSSES. DETERMINATION OF STRESSES. 



31 



TABLE OF STRESSES. 
Compression + ; Tension — . 



Member. 


Dead Load. 


Snow. 


"Wind. 


Max. 


MlN. 


i6 = bs 


+ 43820 


+ 30200 


+ 20170 


+ 94190 


+ 43820 


ft 4 = cu 


+ 37600 


+ 25850 


+ 16550 


+ 80000 


+ 37600 


g2 = dw 


+ 31300 


+ 21570 


+ 12900 


+ 65770 


+ 31300 


fz = ey 


+ 25050 


+ 17250 


+ 10000 


+ 52300 


+ 25050 


6k = rs 


- 40700 


- 28000 


-21770 


- 90470 


- 40700 


5l = qt 


- 40700 


- 28000 


-21770 


- 90470 


- 40700 


3 m = pv 


- 34900 


- 24000 


- 17100 


- 76000 


- 34900 


In = ox 


- 29050 


- 20000 


- 12450 


- 61500 


- 29050 


&6 = st 


- 2720 








- 2720 


- 2720 


34 = uv 


- 5050 


- 1600 


- 1850 


- 8500 


- 5050 


12 = wx 


- 7370 


- 3200 


- 3720 


- 14290 


- 7370 


yz 


- 16700 


- 9600 


- 5570 


- 31870 


- 16700 


45 = tu 


+ 6220 


+ 4320 


+ 5000 


+ 15540 


+ 6220 


23 =vw 


+ 7450 


+ 5100 


+ 5970 


+ 18520 


+ 7450 


zl = xy 


+ 9050 


+ 6220 


+ 7270 


+ 22540 


+ 9050 



Span = 50 ft. Eise of lower chord = 2 ft. Rise of upper chord = 
12 ft. Distance between trusses = 16 ft. Length of middle panel 
of upper chord = 12 ft. The divisions of upper and lower chords 
are otherwise equal. The ventilator roof is raised 4 ft. 

Dead load (total) = 10000 lbs. = 2000 lbs. per panel of upper chord. 

Snow load (15 lbs. per sq. ft. hor. proj.) =2400 lbs. per panel of upper 
chord. 

Wind load = 40 lbs. per sq. ft. on vertical surface = 31 lbs. per sq. ft. on 
slope. 

Total wind load on main slope of roof = 11150 lbs. 

Total wind load on slope of ventilator roof = 3520 lbs. 

Total wind load on vertical surface = 2560 lbs. 

Attention is directed to the following points: 
(1) The ventilator roof, braced as shown by the dotted lines, 
forms a complete truss fixed at the ends. The pressures on the two 



32 NOTES ON GRAPHIC STATICS. 

supporting joints of the main truss, due to loads on the raised portion, 
are then determined in the same manner as are the reactions of a 
truss with fixed ends. Thus the resultant wind pressure on the 
raised portion, for wind on the left side, is a-F, its line of action 
cutting el at x. The resulting pressures, aE and EF, on the two 
joints of the main truss are found by dividing aF into parts inversely 
proportional to the segments of el. The wind pressure on the joint 
cle is then the resultant of aE and Da, the latter being one-half the 
wind pressure on the panel d. 

(2) In finding the supporting forces due to wind pressure by the 
funicular polygon, the resultant of the pressures on the main roof, 
and on the vertical surface of the raised portion, was used in order 
to obtain better intersections than otherwise. 

(3) The snow load is one-fifth greater than the dead load. Since 
both are assumed to be distributed in the same manner, the snow 
load stresses can be found by multiplying the corresponding dead 
load stresses by six-fifths, no snow load diagram being necessary. 

(4) The maximum and minimum stresses are found in a manner 
similar to those of Example 1, with the exception that both sets of 
wind stresses must be considered. The results are given in Fig. 2 d. 
The stresses in kl and Im change sign. It will be seen from the 
diagrams that wind on the fixed side causes compression in kl and 
tension in Im, while the reverse stresses result from wind on the 
other side. 

(5) The stresses in the members of the ventilator roof truss can 
be found by treating it as a truss with fixed ends. 

The student should trace out the construction of the diagrams 
and verify the maximum and minimum stresses. 

40. Trusses having only Two Unknown Forces to determine at 
Each Joint. In the preceding examples only two unknown stresses 
were encountered at each joint, so that the polygons of forces for 
the successive joints of the truss could be constructed at once. 
Other cases of the same nature are given in Fig. 25 A, B, C, D, E, 
F, G, and H. 

41. Fink or French Truss. (See Fig. 26.) This is a form of iron 
roof truss in common use for shops and similar buildings. In con- 
structing the force polygons for this truss, a difficulty is encountered 
to which attention is directed. Beginning at either (the right-hand) 
support, the force polygons can be constructed in the usual manner 
until the joint ruvk is reached, when three unknown forces are 



ROOF TRUSSES. DETERMINATION OF STRESSES. 33 




c'' 




Fig. 25. 



34 



NOTES ON GRAPHIC STATICS. 



encountered. To overcome this difficulty, the stress in rk must be 
found by some other method. 

The most satisfactory solution is to calculate the stress in rk by 
moments, as explained in Art. 6, and insert this calculated stress in 




Wind Right. 



Fig. 26. 



the stress diagram. Then the stresses in the two remaining mem- 
bers, ru and uv f can be found by completing the force polygon. 
A general graphical solution is shown in Fig. 9. (See Art. 14.) 
A simpler graphical solution, shown by dotted lines in the stress 
diagrams of Pig. 26, is as follows : Since, from the solution by 



ROOF TRUSSES. COTTNTERBRACING. 35 

moments, the stress in rk depends only upon the moments of the 
external forces acting on either half of the truss, about the apex of 
the truss as moment axis, a single load acting at the end joint can 
be substituted for the given loading on that side provided that the 
moment about the apex is not altered. When the load is uniformly 
distributed over the rafter, this single load will evidently be equal 
to one-half the load on the rafter. Under this altered loading the 
stresses in the dotted members will be zero, since they serve only 
to support loads at the intermediate joints of the rafters. The 
rafter may then be treated as a single member, pa. Referring to 
the dead load stresses, the forces acting at the right end joint under 
these conditions are /3a, aj = \ W, jk, and k(3. The forces k(3 and 
(3a are then determined by completing the force polygon aJK/3a. 
Next, the force polygon for the joint (3kr is completed, thus deter- 
mining the true stress KB, and locating R in the stress diagram. 

In the wind diagrams this construction is applicable to the 
unloaded side without altering the distribution of the loads, since 
the dotted members on that side are not stressed, for the reason 
stated above. 

It is to be noted that this method is only applicable when the 
two halves of the upper chord are straight lines, hence it could not 
be employed for such a truss as Fig. 25 M. 

Other forms of trusses requiring the same special treatment as 
Fig. 26 are shown in Fig. 25 I, J, K, L, and M. 

§ 3. Counterbracing. 

42. Definitions. It will be noticed that the members of a com- 
plete truss form a system of triangles. The triangle is the element- 
ary truss. Under the action of forces lying in its plane and acting 
at its vertices, it cannot be distorted without changing the lengths 
of one or more of its sides, and such changes are opposed by the 
resistance to extension and compression of the sides. A polygon of 
more than three sides, assumed free to turn at the joints, can be dis- 
torted without altering the lengths of any of the sides. 

The rectangular frame (Fig. 27), acted on by 

the force F, would be distorted as shown, the \^ p 

distance AB becoming shorter and CD longer, 
change in length of the diagonals necessarily 
accompanying the distortion of the frame. A 
diagonal member capable of resisting both exten- 
sion and compression would, then, prevent dis- fig. 27. 



36 NOTES ON GRAPHIC STATICS. 

tortion. A member joining C and D, capable of resisting tension 
alone, would prevent distortion in the direction shown in Fig. 27, 
but not in the opposite direction. Two tension diagonals, however, 
would evidently make the frame stable. Thus a rectangular frame 
may be made capable of resisting any forces acting at the angles, 
tending to distort it, by the introduction of a single diagonal mem- 
ber capable of resisting both tension and compression, or of two 
diagonals, both capable of resisting tension alone or compression 
alone. In the case of two diagonals it is evident that only one of 
them would be stressed at a time. Considering the quadrilateral of 
Fig. 27 to represent one panel' of a truss, the diagonal which is 
stressed, under the action of the dead load is called the main brace; 
the other, the counterbrace or counter. The counter may be stressed 
under the action of the wind, snow on one side of the roof, or other 
non-symmetrical temporary load. 

43. Determination of Stresses. The work of determining the 
stresses in trusses having counterbracing is more involved than is 
the case with the class of trusses previously considered. The basis 
of the treatment is the fact that only one diagonal of a panel is 
stressed at a time. Two methods of solution will be illustrated in 
the following example, each possessing certain advantages. 

Example. The chords of the Crescent truss (Plate II) are circu- 
lar arcs. The left end is on rollers. Span = 80 ft. Rise of upper 
chord = 26 ft. Rise of lower chord = 10 ft. Distance between 
trusses = 20 ft. The truss is divided by verticals into 5 panels, each 
16 ft. wide. The diagonals are to be tension members. Assumed 
weight of truss = 10800 lbs. Weight of roof surface, etc. = 20 lbs. 
per sq. ft. Snow load = 20 lbs. per sq. ft. of horizontal projection. 
Wind load = 40 lbs. per sq. ft. on a vertical surface. The wind 
pressure on each panel is assumed to act at right angles to the chord 
of the arc. The joint loads used are given in the following table. 



ROOF TRUSSES. COUNTERBRACING. 



37 



Table of Joint Loads. 



Joint. 


Truss. 


EOOF, ETC. 


Total 
Dead Load. 


Snow. 


Wind K. 


Wind L. 


AB 
BC 
CD 
BE 
EF 

Fa 


lbs. 
1080 

2160 

2160 

2160 

2160 

1080 


4940 
8390 
6650 
6650 
8390 
4940 


6020 
10550 

8810 

8810 
10550 

6020 


3200 
6400 
6400 
6400 
6400 
3200 


3450 

( 3450 
\9390 

9390 


9390 

f 9390 
\3450 

3450 



44. Lettering Truss Diagram. An adaptation of Bow's notation 
is used. One diagonal of each panel is dotted. The usual method 
of lettering is followed, with the exception of the panels having coun- 
terbracing, e.g. mn represents the diagonal of the second panel, which 
is drawn in full lines, while the same letters accented will be used to 
represent the dotted diagonal. While we are dealing with either 
diagonal of any panel, the other is considered to be removed, since 
only one of them is stressed at a time. If we are dealing with mn 
and l'k', for example, the vertical will be represented by nV ; the seg- 
ments of the upper chord will be cn and dV ; the segments of the 
lower chord, mr and k'r. Again, when dealing with the diagonals 
m'n' and l'k' the vertical between them is m'V ; the segments of the 
upper chord are cn' and dV, and of the lower chord m'r and k'r. 
Thus the letters which represent the verticals and chord segments 
vary, depending upon which diagonals are under consideration. 

45. Construction of Diagrams. First Method. By this method 
the diagrams are constructed for each kind of loading separately (see 
Figs, la, lb, lc, Id, Plate II). The line of action of the result- 
ant load in each case passes through the centre of curvature (O) of 
the upper chord, and is parallel to the closing side of the force poly- 
gon. The reactions for the wind diagrams are found from the 
condition that the resultant load and reactions must intersect at 
a common point. The construction is not shown. 

In determining the stresses in the truss members, one set of diago- 
nals (e.g. those dotted) are omitted, and the force polygons for all 
the joints are drawn in the usual way. The omitted set of diagonals 
are then dealt with, and the construction repeated. This second 



38 



NOTES ON GRAPHIC STATICS. 



part of the construction only requires the drawing of the three 
additional lines corresponding to the three diagonals omitted in the 
first instance. The stress diagram thus contains the stresses in all 
the members of the truss, under the assumption that either diagonal 
in each panel is the one stressed. E.g. in the "wind right" dia- 
gram NL' is the stress in the vertical when the diagonals nm and 
I'k' of the two adjacent panels are assumed to be in action, while 
M'K is the stress in the same member when the diagonals in action 
are n'm' and Ik. The student should trace out the construction of 
the diagrams. 

46. Determination of Maximum Stresses. The stresses are scaled 
off and inserted in the table (p. 39). The following combinations 
of loads will be considered : (1) dead load ; (2) dead and snow loads ; 
(3) dead load and wind on right side; (4) dead load and wind on 
left side ; (5) dead and snow loads with wind right ; (6) dead and 
snow loads with wind left. The basis for the determination of the 
stresses due to these various combinations of loads is the fact, that, 
since the diagonals are to be tension members, the one will be 
stressed which is in tension under the given set of loads. From the 
tabulated diagonal stresses we find that the diagonals in action are 
those given in the following table : 



Diagonals in Action. 



Dead. 


D. & 8. 


D. & W. E. 


D. & W. L. 


D. S. &W. E. 


D. S. & W. L. 


mn 
neither 


mn 
neither 


mn 

Ik 

j'i' 


m'n' 
I'k' 


mn 

Ik 

j'i' 


m'n' 

I'k' 

ji 



We can now determine the actual stress in any member under 
any combination of loads. Thus to find the stress in the second 
vertical under the action of D.S. & W.K,., we see from the table 
that the diagonals in action in the adjacent panels are mn and Ik. 
The stress in the vertical is then represented by NK. Taking the 
algebraic sum of these stresses as given in the line nk of the table 
(p. 39), we find the required stress to be - 2760 - 1520 + 1000 = 
— 3280 lbs. The stress in this member for each combination of 
loads is found in a similar manner. The extreme results are 
recorded in columns 6 and 7 of the table. The student should 
determine the maximum and minimum stresses in all the members 
of the truss. 



ROOF TRUSSES. COUNTERBRACING. 
Table of Stresses. 



39 



Mem- 



mn 
in'n' 

Ik 
l'k' 

ji 
j'i' 

bo 
( en 
[ en' 
(dl 
\dV 

{ej 
( ej' 

fh 

or 

{mr 
m'r 
Ikr 
\k'r 
( ir 



Dead. 



i'r 
hr 
om 
on' 
nk 
m'V 

W 
[k'j 
\ih 
f j'h 



- 4180 
+ 5600 




- 4180 
+ 5600 
+38980 
+31080 
+27180 
+28830 
+28830 
+31080 
+27180 
+38980 
-27250 
—25650 
-29370 
—28820 
-28820 
-25650 
-29370 
—27250 

- 5500 

- 9000 

- 2760 

- 5550 

- 2750 

- 5550 

- 2760 

- 5500 

- 9000 



Snow. 



— 3450 
+ 4600 




— 3450 
+ 4600 
+25720 
+21200 
+17970 
+19650 
+19650 
+21200 
+17970 
+25720 
-17980 
-16950 
—20000 
-19660 
-19660 
-16950 
-20000 
-17980 

— 3600 

— 6550 

— 1520 

— 3800 

— 1520 

— 3800 

— 1520 

— 3600 

— 6550 



Wind K. 



- 4800 
+ 6370 

- 7970 
+ 7970 
+ 8060 
-10730 
+11050 
+12200 
+ 7700 
+17030 
+11310 
+16960 
+24450 
+22000 

- 7690 

- 7250 
-11490 
-11320 
-17020 
-24430 
—17330 
-25920 

- 1530 

- 5580 
+ 1000 

- 7710 

- 3100 
+ 2270 

- 8670 

- 5200 
+ 1600 



"Wind L. 


Max. 




(12450) 


+ 9690 


—12430 




(7300) 


-12910 


— 7310 




(8000) 


+ 7920 


— 7970 




(7850) 


- 7920 


— 7920 




(10700) 


- 3150 


—10780 




(5150) 


+ 4200 


— 5130 




(75650) 


+ 8070 


+75750 


+ 5670 


(64350) 




+64480 


+14700 




+ 870 


(65500) 




+65510 


+ 6540 




+ 930 


(69450) 




+69600 


— 2030 






(86650) 


- 2840 


+86700 




(52900) 


+ 2110 


—52920 


+ 2000 


(49900) 




—49850 


+10570 




+10360 


(59700) 




—59800 


+16040 




+19130 


(66700) 




—66700 


+16330 






(71050) 


+20250 


—71150 


+ 400 


(10500) 




—10630 


+ 8580 




— 4440 


(4250) 




- 4280 


+ 7500 




- 360 






(7100) 


— 2440 


- 7080 


+ 5160 




+ 4070 


(14000) 




—13950 


+ 1400 





MlN. 









+38980 

+31080 
+28830 

+31080 

(36200) 
+36140 

(25100) 
—25140 

(18750) 
—18800 



(12850) 
-12780 



(6650) 

• 6520 

(7100) 

• 7000 

(400) 
420 



(1950) 
+ 1950 



(2300) 
+ 2400 



(1350) 
1430 



Loads Caus- 
ing Max. S. 



D.S.&W.R. 

D.&W.L. 

W.R. 

W.L. 

D.S.&W.L, 

D.&W.R. 

D.S.&W.R 

D.S.&W.R 
D.S.&W.R 

D.S.&W.R 

D.S.&W.R 
D.S.&W.R 

D.S.&W.R 

D.S.&W.R 

D.S.&W.R 
D.S.&W.R. 
D.S.&W.R. 

D.&S. 

D.S.&W.R. 

D.S.&W.R. 



40 NOTES ON GRAPHIC STATICS. 

47. Construction of Diagrams. Second Metliod. Thus far the 
stresses due to wind, snow, etc., have been determined separately. 
Another method is to draw the stress diagram for each combination 
of loads, in which case the maximum and minimum stresses can be 
taken at once from the diagrams. The latter method possesses 
greater advantages when dealing with trusses having counterbracing 
than otherwise. In the example, six diagrams would be required 
by the second method, as six combinations of loads are considered. 
The dead load diagram is not repeated. The remaining ones are 
given in Figs. 1 e, If, lg, lh of Plate II. The same lettering is 
used as for the first method. Referring to Fig. le, the external 
forces are laid off in order. At the second joint, for example, the 
resultant load is BC, made up of Ba, which is one-half the wind 
load on b ; aft, one-half the wind load on c; /3(7, the vertical load 
on the joint. The line of action of the resultant load, marked 
D. & W., acts through the centre of curvature and parallel to AF. 
The reactions FR and RA are found by applying the funicular 
polygon to the resultant load. 

In constructing the force polygons for the joints, that diagonal 
of each panel which is found by trial to be in tension under the 
given loads is used. It is of advantage to draw all the lines corre- 
sponding to the chord stresses at first. Then (Fig. 1 e), having 
constructed the polygon ABOR, we must determine which diagonal 
of the second panel is in action. Omitting the dotted one, we con- 
struct the polygons OMR and BCNMO. The order of letters indi- 
cates that mn would be in compression, and hence that m V is the 
diagonal in action. The final construction is then made, using the 
latter diagonal. It is not necessary actually to draw the additional 
lines of the trial polygons. OM must be drawn, and its intersection 
with MR locates M. CN being already drawn, the position of M 
relatively to OiVwill indicate the kind of stress in mn. Each panel 
is dealt with in a similar manner. The construction of the remain- 
ing diagrams is like that explained. 

Figure 1 h contains the stresses due to D.S. & W.B. and also D.S. & 
W.L. This combination diagram is made to save space and labor. 
It is evident from the symmetry of the truss that the stress in any 
member due to D.S. & W.L. is the same as that which would exist 
in the corresponding member on the other side, if the wind acted on 
the right side and the roller were transferred to the right end. By 
following this course the two wind diagrams can be combined as 
shown. The additional construction lines are dotted and letters 
underscored. For example, MR is the stress in ir, MN in ij, etc., 
for D.S. & W.L. loads, the roller being at the left end. 



DOUBLE SYSTEM OF WEB MEMBERS. 



41 



48. Maximum and Minimum Stresses. The maximum and mini- 
mum stresses can be determined at once by comparing the lengths 
of the corresponding lines in the various diagrams. The results, 
obtained independently of those previously given, are inserted in 
parentheses in columns 6 and 7 of the table (p. 39). The student 
should trace out the construction of the diagrams, and determine 
the maximum and minimum stresses from them. 

Each of the methods described possesses advantages. The deter- 
mination of maximum and minimum stresses is more laborious by 
the first method; on the other hand, it is desirable to know the 
stresses produced by the separate kinds of loads. The first method 
is favorable to the graphical work, as the diagrams are smaller or 
may be drawn to a larger scale, and are likely to be fewer in number. 
(See also Art. 28.) 

Figure 25 N and are other examples of trusses with counter- 
bracing. 



§ 4. Trusses ivith a Double System of Web Members. 



Such a truss may be treated as 
b 



49. General Methods of Solution. 

a combination of two trusses having 
common chords but distinct sys- 
tems of web members. The girder 
of Fig. 28 A, for example, can be 
resolved into those shown in Figs. ' 
28 B and 28 C, the stresses in 
the members of these component 
trusses being found in the usual 
manner. The actual stress in any - 
web member of the original truss 
is given directly by the diagrams, 
while the stress in any chord seg- 
ment, as ab, is evidently equal to 
the algebraic sum of the stresses 
found for cd (Fig. 28 B) and ef 
(Fig. 28 C). When the given truss can be resolved in more than one 
way, or when the distribution of loads between the component 
trusses is uncertain, the problem is indeterminate. 

Figure 29 is a Crescent roof truss with two systems of web mem- 
bers, as shown by the full and dotted lines respectively. By tracing 
out the force polygons for the different joints it will be seen that the 
stress diagram can be drawn at once for the complete truss. A 




42 NOTES ON GRAPHIC STATICS. 

suitable system of lettering the interior of the truss is given, the 
intersections of the diagonals being treated as joints. If the joint 




Fig. 29. 

1 were made to coincide with 2, the portion of its load supported 
by each component truss would be uncertain and the problem in this 
particular would be indeterminate. 

50. Double Diagonal Bracing. In the case of counterbraced 
panels (Fig. 1, Plate II; Fig. 25 N and 0), the two diagonals are 
assumed not to be in action for the same loading, both being designed 
for the same kind of stress. When the diagonals are designed to act 
simultaneously (one being in tension while the other is in compres- 
sion), the truss can be resolved into two trusses having common 
chords and verticals but distinct systems of diagonals. Each joint 
would belong to both component trusses, and the division of its load 
between the two trusses would be uncertain. One way of dealing 
with this case is to assume that the load at each joint is divided 
equally between the two trusses. The stresses in the verticals and 
chords would be the algebraic sum of those found for the separate 
trusses. The results of this method are evidently liable to err in 
the wrong direction. 

§ 5. Three-hinged Arch. 

51. Definition. A three-hinged arch consists of two arched ribs 
hinged at the crown and abutments (Fig. 30). The outward thrust 
at the ends may be resisted by the abutments or by a tie rod joining 
the ends. The ribs may be braced or solid. 

52. Determination of Reactions of Hinges. The reactions are 
assumed to act through the centres of the hinges. The reactions of 
the hinge at the crown on the two half-ribs must evidently be equal 
and opposite. Let ab (Fig. 30) represent the line of action of the 



THREE-HINGED ARCH. 



43 



resultant load supported by the left half-rib, the right half being 
assumed to be unloaded. The reactions at 0' and 0" must be equal 
and act along the line 0"0'. The three forces acting on the left 
half must, for equilibrium, intersect at a common point. This con- 
dition determines on to be the direction of the reaction at 0. 




Fig. 30. 



The magnitudes of these reactions can now be found by construct- 
ing the triangle of forces ABD'. BD' is the magnitude of the 
reaction at 0", and of the two equal and opposite reactions at 0'. 
DA is the magnitude of the reaction at 0. 

The directions and magnitudes of the reactions at 0, 0', and 0", 
for any resultant load be on the right half-rib, are found in a similar 
manner, the triangle of forces being BCD". When both sides are 
loaded, the reaction at either hinge is evidently the resultant of the 
reactions due to the loads taken separately. Combining the two 
separate reactions for each of the hinges by the triangle of forces, 
we find the resultant reactions for the right half-rib to be CD and 
DB at 0" and 0' respectively ; and for the left half -rib, BD and 
DA at 0' and respectively. 

The lines of action of these reactions are ox, xo'y, yo", drawn 
parallel to DA, DB, and DC, respectively. These lines should 
intersect on ab and be as shown, if the construction is accurate. 

In the case of a symmetrical arch, symmetrically loaded, the 
reactions at the crown will evidently be horizontal. The reactions 
of the end hinges can then for this case be found directly; the 
points x and y being determined by the intersections of a horizontal 
line through 0' with the resultant loads. 



44 



NOTES ON GRAPHIC STATICS. 



53. Determination of Reactions. Second Method. The point D 
(Fig. 30) can be taken to be the pole, and the lines ox, xy, yo" the 
strings of a funicular polygon for the given loads. The reactions 
of the hinges can, then, be determined by constructing a funicular 
polygon for the loads, such that the three strings will pass through 
the hinges. The strings will be the lines of action of the reactions, 
and the lengths of the corresponding rays their magnitudes. 

Instead of using the resultant loads, the actual loads may be 
employed in applying this method. In this case it is only necessary 
to construct a funicular polygon for the given loads, such that the 
three limiting strings will pass through the three hinges. (For 
methods, see Arts. 23 and 24.) 

54. Determination of Reactions. Algebraic Solution. In order to 
secure greater accuracy, it may be desired to determine by calcula- 
tion the position of the pole of the required funicular polygon. 
This is done by calculating the reaction at 0', since the ray DB 
(Fig. 30) or PD (Fig. 31) represents this reaction. 

Let B, R be the equal and opposite reactions at 0' (Fig. 31). 
These are resolved into horizontal (H) and vertical ( V) components. 




Fig. 81. 



Taking the moments, about 0, of the forces acting on the left half, 
we have H • b — V • a = 2 W'x', in which W represents any load, and 
x' its arm. Similarly, the moments about 0" of the forces acting on 




THREE-HINGED ARCH. 45 

the right half give the equation H • b + V • a = 2 Wx. By solving 
these two equations, we determine the values of H and V, and con- 
sequently R. This value of R, laid off in the proper direction from 
the point D of the force polygon, locates the pole P of the required 
funicular polygon. PA and PG are then the reactions at and 0" 
respectively. It is to be noted that PAD and PDG are the poly- 
gons of external forces for the two half-ribs. 

Problem. The semicircular arch (Fig. 32), hinged as shown, is 
loaded with a dead load of 8000 
pounds uniformly distributed 
over the roof, and also with a 
wind load of 8000 pounds on 
the right side. Find the reac- 
tions of the hinges : (1) graphi- 
cally ; (2) by calculation, as 
explained. Also draw a funic- 
ular polygon for these loads, to 
pass through the three hinges. 

55. Determination of Stresses in Braced Arches. Method of Joints. 

When the two half-ribs are braced, as shown in Fig. 3, Plate III, the 
stresses in the members can be found by constructing the polygons 
of forces for the joints in succession, the methods being the same 
as those explained in the preceding sections of this chapter. When 
the number of panels is large, the treatment shown in Fig. 37, 
Art. 62, should be employed in order to avoid the inaccuracies result- 
ing from a long succession of joints, and to check the work. 

56. Example. (See Fig. 3, Plate III.) The general dimensions 
of the arch are given in the drawing. The inner curve is a semi- 
circle of 40 ft. radius. The outer curve consists of two circular 
arcs of 65 ft. radius, intersecting at the apex. The dead load is 
taken to be 3000 lbs. per panel. In estimating the wind load, the 
panels were taken in pairs, the wind being assumed to act at right 
angles to the chord extending across two panels. The total wind 
pressure on the panels g' and/' is 4400 lbs. ; on e' and d', 9000 lbs. ; 
and on c' and b', 14400 lbs. From the symmetry of the arch, the 
maximum stresses in corresponding members on the two sides will be 
equal, and it is only necessary to construct the dead load diagram 
for one side, and the wind diagram for wind pressure on one side. 
The snow stress diagram is not drawn. 

. Dead Load Diagram. The loads on the right half -arch are plotted 
to scale, GG' being one-half the load at the crown joint. Since the 



46 NOTES ON GRAPHIC STATICS. 

reaction of the crown hinge is horizontal (Art. 52), the pole P' is 
taken on a horizontal line through G, and the dotted funicular poly- 
gon (2') is constructed, the first string being drawn parallel to P'A' 
through the right-hand hinge. The pole P of the polygon which 
will pass through both hinges is now located by the method of 
Art. 20. G2 and G2' are laid off equal, respectively, to the ordi- 
nates of polygons (2) and (2') measured from /?. Connecting 2 with 
P', and drawing a parallel through 2', P is located, in accordance 
with the condition of Art. 20. The required polygon (2) is then 
drawn, working from both hinges and closing half-way between the 
hinges. Otherwise the angles of this polygon could be located 
directly by the second method of Art. 24. 

The reactions of the hinges being A'P and PG, the polygon of 
external forces for the right half-arch is PGG'F'E'D'C'B'A'P. 
The stresses in the truss members can now be determined by con- 
structing the polygons of forces for the joints in succession, begin- 
ning at the hinges. To insure greater accuracy and also check the 
work, the stress in the member n'p of the lower chord was calculated 
by moments. This may be done as follows : 

The string of the funicular polygon, corresponding to any panel 
of the truss as d', is the line of action of the resultant of all the 
external forces (PG, GG', G'F', F'E', E'D') acting on either side 
(left) of the panel ; hence it is the line of action of the resultant 
external force acting on the members cut by the section x'y', the 
magnitude of this force being the corresponding ray PD'. This 
force PD' is balanced by the stresses in the members d'm', m'n', and 
n'p. The stress in n'p can then be found by taking moments about 
the intersection of d'm' and m'n', i.e. the moment of the external 
force PD' about this axis is balanced by the moment of the stress in 
n'p. By measurement, PD' = 2".56 x 5000 = 12800 lbs. and its arm 
is 1".06 on the drawing. The arm of the chord n'p is .94"; hence, 

a , • , 12800x1.06 14mi , " / . s 

Stress m n'p = = 14430 lbs. (compression). 

F .94 \ v ) 

This stress is plotted in proper position in the stress diagram, and 
the point N' is thus located. The stresses in d'm' and m'n' are now 
determined by completing the polygon of forces PGG' F'E' D'M' N'P. 
The polygons of forces at the different joints of the arch can now be 
constructed in the usual way, working from the hinges and also from 
the panel d' in both directions. 

Wind Diagram. The wind loads are plotted to scale in Fig. 3 b, 
and their lines of action are then drawn parallel to the proper sides 



THREE-HINGED ABCH. 47 

of the force polygon. Assuming any pole P', a funicular polygon 
zz' for these loads is constructed. 

The pole P of the polygon to pass through the three hinges is 
located as follows : Since the left side is unloaded the reaction of the 
crown hinge must also pass through the left-hand hinge (Art. 52). 
The corresponding ray PG is drawn through G parallel to this reac- 
tion, and the required pole must lie on this ray. The pole must also 
lie on Pd, this line being located by the method of Art. 23 (the 
construction is shown by dotted lines). The intersection of these 
two lines PG and Pd locates the pole P. The funicular polygon 
marked (1) is constructed from this pole, the strings PG and PA' 
passing through the hinges as required. The polygon of external 
forces for the whole arch is PGG'F'E'D'C'B'A'P. Having deter- 
mined the hinge reactions, the polygons of forces at the joints can 
be constructed. 

To increase the accuracy of the diagram, the stresses in pn' and 
pn were calculated in the manner previously described for the dead 
load. Keferring to pn, the external force (reaction of left hinge) is 
found to be 2".72 x 5000 = 13600 lbs. Arm 3-4 = 1.87", arm 3-5 
= .94". Taking moments about the joint (3), we have 

a , • 13600x1.87 oTinmu 
Stress in np = = 27100 lbs. 

* .94 

This stress is compression. By the same method we have 

Stress in n'p = 54Q0 X 3 - 84 = 22100 lbs. (tension). 
1 .94 v J 

These stresses were plotted in Fig. 3 b ; then the force polygons for 
the joints were constructed as explained for Fig. 3 a. The student 
should trace out the construction. 

57. Maximum Stresses. The maximum stress in any member, as 
d'm' = dm, is found as follows : 

The dead load stress, scaled from the diagram, is 

.34 x 5000 = 1700 lbs. (tension). 

The stress in m'd' for wind right is 5.53 x 5000 = 27650 lbs. (com- 
pression) = stress in md for wind left. The stress in m'd' for wind 
left is the same as that in md for wind right, and equals 

2.94 x 5000 = 14700 lbs. (tension). 



48 NOTES ON GRAPHIC STATICS. 

Comparing these results, the extreme stresses in m'd' and md are 

found to be 

14700 + 1700 = 16400 lbs. (tension) 

and 27650 - 1700 = 25950 lbs. (compression). 

The funicular polygons and stress diagrams might have been con- 
structed for the combined loadings, as explained for simple trusses. 
Counterbracing can also be dealt with in the same manner as 
previously explained. 

The funicular polygon drawn through the hinges is called the 
line of pressure, since its strings represent the lines of action of the 
resultant pressure on the panels corresponding to these strings. 




Fig. 



Figure 33 represents one of the arches supporting the train-shed 
roof of the Philadelphia and Reading Terminal Railway. The 
diagonals are tension members. 1 

58. Three-hinged Arch. Solid Ribs. Determination of Stresses. 

Let Fig. 34 represent such an arch, the funicular and force polygons 
for the given system of loads being drawn. The resultant external 
force acting at a cross section n is PA = F, its line of action being 
the corresponding string. This force can be resolved into F' and F", 
acting respectively at right angles and parallel to the plane of the 
section. The latter is the shearing force at the section. The former 
(compounded with two equal and opposite forces F' acting through 
n, the centre of gravity of the section) is equivalent to a force F' 
acting at the centre of gravity and causing a uniformly distributed 
compression stress, and a couple whose arm is nn'. This couple 
causes, as in a beam, a bending stress at the section, the bending 
moment being F' x nn', or its equal F x ny. The bending stress 

1 Yot description, see Trans. Am. Soc. C, 2?., 1895. 



TRUSSES LOADED AT OTHER POINTS THAN THE JOINTS. 49 

is combined with the direct compression stress, as in case of a strut 
loaded eccentrically. The stress at any other section is found in a 
similar way. 




The determination of the stresses in arches fixed at the ends, 
and in arches hinged only at the ends, requires the use of the theory 
of elasticity. For the treatment of such cases, see the list of refer- 
ences, p. 77. 



§ 6. Trusses loaded at Other Points than the Joints. Bending Stresses. 

59. Determination of Joint Loads. The purlins are frequently 
supported at other points than the joints. The design of the roof 
may require the purlins to be placed closer together than the joints 
of the upper chord, e.g. when the roof covering, as slate or corrugated 
iron, is supported directly by the purlins. In such cases the upper 
chord is subjected to a bending as well as a direct stress, and this 
must be provided for in proportioning this member. The lower 
chord also may be subjected to bending stresses from shafting or 
other loads supported by it at other points than the joints, as well 
as from its own weight. 

Example. Let the total weight of roof covering and purlins 
(Fig. 35 A) be 3000 lbs. The weights at the points of support of 
the purlins will then be those given. Consider the load at n. (See 
Fig. 35 B.) This load can be resolved into 167 lbs. at A and 333 lbs. 
at B. If, now, we apply at A and B equal and opposite forces equal 
to the components, these will balance and not affect the stresses 
in the truss. The upward forces at A and B correspond to the 



50 



NOTES ON GRAPHIC STATICS. 



reactions of a beam under the load of 500 lbs., and there remain the 
loads, 167 lbs. at A and 333 lbs. at B. This 500 lb. load is, there- 
fore, equivalent to the component loads at the adjacent joints, and 
in addition causes a bending stress in AB, the maximum bending 
moment being at the load and equal to 167 x 5 ft. lbs. The joint 




loads found as indicated are given in Fig. 35 A. All other loads 
supported by the purlins are treated similarly. This method is 
evidently approximate when the chord is continuous, as is commonly 
the case, and when the loading is oblique to the length of the chord. 




W 



if 




60. Combined Stresses. In Fig. 36 let the piece be subjected 
to a transverse load W and a direct force P, acting through the 

centre of gravity of the end sec- 
tions. Let the resulting deflection 
be v. We have then : (1) a bend- 
ing stress due to W; (2) a bending 
stress due to the eccentricity v of 
P; (3) a direct compression stress. 
When the deflection v is sufficiently small, the bending stress (2) 
can be neglected, as is commonly done. In this case the resultant 
stress is the algebraic sum of the direct stress and the bending 

P M 

stress due to W, i.e. /=-r ± — V in which M is the bending moment 

due to W, 



Fig. 36. 



GENERAL DIRECTIONS. 51 

If the force P does not act through the centre of gravity of the 
end sections, there results an additional bending stress due to this 
eccentricity. 

§ 7. General Directions. 

61. Instruments. In order to secure a satisfactory degree of 
accuracy in graphical solutions, special precautions must be taken. 
In general, the drawings should be made with such care and in such 
a manner as to leave no question as to their accuracy at any stage 
of the construction. 

The wooden edges of the ordinary drawing board and T-square 
are unreliable. A steel straight edge with lead weights to hold it 
in position is preferable. The edges of the triangles must be 
straight and the 90° angle true. The usual hard rubber and cel- 
luloid triangles are not sufficiently accurate in these respects and 
should be tested before using. The best quality of dividers and 
compasses must be used. A metal scale graduated to hundredths of 
an inch and a very hard pencil kept sharpened to a fine chisel edge, 
complete the list of necessary instruments. All intersections to be 
preserved are located accurately by a fine prick point enclosed in a 
circle. 

62. Construction of Diagrams. The space diagram is constructed 
accurately and, if the directions of the lines of the stress diagram 
are to be taken from it, to a large scale. In constructing the stress 
diagram the directions of its lines are commonly taken from the 
space diagram. In doing this, a good general rule to follow is not 
to obtain the direction of a line of the stress diagram from a shorter 
line of the truss. In order to secure accuracy in this respect with- 
out constructing the truss to an extravagantly large scale, the 
construction may sometimes be made by indirect means. 

Illustration. The direction of BS (Fig. 1 a, Plate I) may be 
obtained by laying it off to the same slope as the rafter {i.e. making 
ES = j-QrBR), instead of drawing it parallel to that member in 
Fig. 1. CU, DW, etc., are drawn parallel to BS. ST, UV, etc., 
are drawn perpendicular to the base line. TU, VW, etc., are drawn 
to the same slope as the members of the truss. Thus the whole 
stress diagram might be constructed without taking the direction 
of any line from the- truss diagram. When the chords are arcs of 
circles, the chord stresses should be drawn at right angles to the 
radius bisecting the chord member. 

The intersections of the lines of the stress diagram should be 



52 



NOTES ON GRAPHIC STATICS. 



located accurately. If the slope of the rafter is small, the angle 
between it and the lower chord is very acute and the intersection 
of the corresponding lines BS and BS (Fig. 1, Plate I) is liable to 
considerable error. As the accuracy of the remainder of the work 
depends upon the correct location of the point S, it may be desirable 
to locate it by calculation in some such way as explained in Art. 36. 

In important work it is well to determine the supporting forces 
by calculation, as the correctness of all the stresses depends on 
them. All such calculations serve as checks on the graphical 
work. 

As the construction proceeds from joint to joint of the truss, 
errors accumulate ; hence a long succession of joints in construct- 
ing the stress diagram should be avoided. In a truss of compara- 
tively few panels it is sufficient to work from each end towards the 
centre. In case of a large number of panels, the truss should be 
broken up into sections, using the method of sections (Art. 6, Third 
Solution). In Fig. 37, for example, we can find the stress in mn by 
moments and lay off the result JOT" in the stress diagram. By the 




1A 



Fig. 37 



polygon of forces we then determine FO and ON to be the stresses 
in these members. We can now complete the diagram, working 
from each end towards the centre and from the centre towards 
the ends, closing half-way between centre and ends. In the same 
manner the truss may be divided into any number of sections. 



63. Checks. The accuracy of the construction is checked by the 
diagrams closing. This, however, does not insure the correctness 



GENERAL DIRECTIONS. 53 

of the results, as the truss diagram may have been drawn incor- 
rectly or errors made in plotting the loads. It is well, in addition, 
to check any important work by computing the stresses in one or 
more of the members by moments or otherwise. When the compu- 
tations by moments are made at the outset, and the results introduced 
into the diagrams, as indicated in the preceding paragraph, the proper 
closing of the diagrams is equivalent to both the checks mentioned. 

64. Scale of Stress Diagram. Nothing is gained by making this 
scale excessively large. In small trusses it can always be made 
large enough without making unwieldy diagrams. The lines of the 
stress diagrams, when drawn with the precautions described above, 
need not be in error more than two or three hundredths of an inch. 
With a scale of 10000 lbs. to the inch, this would be an error of 
200 to 300 lbs. in the stress. Such an error is of no importance in 
large work requiring so small a scale. 

The student is recommended to study trusses with the view of under- 
standing the purpose and action of their various members independently 
of the stress diagrams. In Fig. 24, for example, the load cd tends 
to deflect the rafter, this deflection being prevented by the brace kj, 
which would therefore be in compression, kj, being in compression, 
exerts a downward thrust on the lower chord at a. The lower chord 
is prevented from deflecting under this thrust by ji ; ji is therefore 
in tension. A study of trusses in such a manner will assist the 
student to a better understanding of their design. 



CHAPTER III. 



BEAMS. 



§ 1. Shearing Force and Bending Moment. 

65. Definitions. The shearing force at any section of a beam is 
equal to the algebraic sum of the external forces to the left of the 
section. 

The bending moment at any section of a beam is equal to the 
algebraic sum of the moments of the external forces to the left of 
the section, the moment axis being the neutral axis of the section. 

It was shown in Art. 18 that if an ordinate be drawn at any 
section of a beam, the distance intercepted on this ordinate by the 
funicular polygon, multiplied by the pole distance, is equal to the 
bendiug moment at the section. The scale of the ordinate is 
the same as that of the space diagram, and the scale of the pole 
distance is the same as that of the force polygon. 

66. Beam supported at Ends and loaded with Concentrated Loads. 

The force and funicular polygons (Fig. 38) are drawn, and the sup- 




Fig. 38. 



porting forces DE and EA determined by means of them. (See 
Art. 12.) By Art. 18, the bending moment at any section, as mn, 

54 



SHEARING FORCE AND BENDING MOMENT. 



55 



is equal to the intercept mn, times the pole distance. The intercept 
raV, in the lower diagram, represents the shearing force at the sec- 
tion. This shear diagram is constructed directly from the definition 
of shearing force, and needs no description. It will be seen by 
inspection of these two diagrams that the maximum shearing force 
is at one end and the maximum bending moment at one of the 
loads, be. 

67. Overhanging Beam. (Fig. 39.) The construction will be 
understood without explanation. The shear diagram is omitted. It 



8- 



C- 

0- 




Fig. 39. 



will be noticed that the bending moments at x and x' are each zero. 
These points are therefore the points of inflexion of the elastic 
curve which the beam will assume under the given loads. The 
bending moment between x and x' is positive, and the elastic curve is 
convex downward. Outside of these points the moment is negative, 
and the curve is consequently convex upward. 

68. Distributed Loads. (Fig. 40.) Let the load be divided as 
shown by the broken lines, and each division be treated as concen- 
trated at its centre of gravity. These concentrated loads can be 
dealt with as indicated in the preceding cases. The funicular poly- 
gon for them is drawn. The bending moments at the points of 
division of the load are the same as those caused by the concen- 
trated loads, and are therefore correctly given by the intercepts of 
the funicular polygon. The exact diagram for the loaded portion 
of the beam is then the inscribed curve, the points of tangency, cor- 
responding to the points of division of the load, being indicated by 
circles. For a uniformly distributed load, this curve is a parabola. 
The shear diagram needs no explanation. 

If the bending moment at any given section of the beam is 
desired, it can be found by making this section a point of division 



m 



NOTES ON GRAPHIC STATICS. 



of the load. Thus the ordinate nn' (Fig. 41) represents the bending 
moment at that section for a load uniformly distributed over the 




Fig. 40. 



entire beam, the funicular polygon being constructed by dealing 
with the loads on the two sides of this section as concentrated at 




JO- 







Fig. 41. 

their centres of gravity. The true diagram is the parabola drawn 
in broken lines tangent to the strings b, a, and c, at %' and the ends 
of the beam. 

69. Problems. 1. Assume a beam loaded with given concen- 
trated loads. Find the bending moment at any section and check 
the result by calculation. Also determine the maximum bending 
moment. 

2. Assume a beam loaded with a given uniformly distributed 
load extending over part of the span. Find the bending moment at 
any section and check the result by calculation. 

3. In Prob. 2 draw the moment curve and determine the value 
of the maximum bending moment. Check the result by calculation. 



DEFLECTION OF BEAMS. 



57 



§ 2. Deflection of Beams. 

70. Graphical Determination of Elastic Curve. In simple cases 
the usual formulas furnish the best method for determining the de- 
flection of a beam. For beams of non-uniform section, or unusual 
system of loads, the deduction of the formulas becomes tedious. In 
such cases a graphical solution can be used to advantage when ex- 
treme accuracy is not required. 

Example. The beam (Fig. 42), of uniform section, is supported 
at the ends and loaded with two concentrated loads as shown. It is 
required to construct the elastic curve. 

Solution. Construct the funicular polygon (Fig. 42 A) for the 
given loads. The force diagram is Fig. 42 A'. Treat the surface of 




Fig. 42. 



this polygon as if it represented a distributed load for the same 
beam, and construct a second funicular polygon. For this purpose 
the surface (Fig. 42 A), is subdivided by ordinates, and each division 
is concentrated at its centre of gravity, marked by a circle. The 
areas of these surfaces are plotted to scale in the force diagram 
(Fig. 42 B'), and the corresponding funicular polygon (Fig. 42 B) is 



58 :notes on graphic statics. 

drawn. The pole P' is taken so that SS 1 is horizontal. The exact 
diagram will be a curve Id scribed in this polygon (Art. 68). This 
curve is the elastic curve of the beam, the deflection at any point 
being represented to a known scale by the length of the intercept 
between the curve and the line SS', this line being drawn so as to 
satisfy the condition that the deflection at each support is zero. 

Proof. Let = any ordinate of Fig. 42 A (in., full size). 
Let D = pole distance of Fig. 42 A' (lbs.). 
Let D' = pole distance of Fig. 42 B' (sq. in., full size). 
Let M= bending moment at any section of the beam 
(in. lbs.). 

Take the origin at the left support, and let C be any section of the 
beam whose distance from the origin is x. 

The ordinate y (Fig. 42 B) at the given section is intercepted by 
SS' and mn, the latter being the tangent to the funicular curve at 
C". Draw the rays P'S and P'C parallel to these lines. The angle 
CP'S is then equal to the angle of slope i of the tangent mn. 

Hence, tan < = SL gg = SA ' ~ A '° . . . . (1) 

dx P'S P'S -=D' W 

In Eq. 1 SA' is constant and A'C is equal to the area in 
Fig. 42 A, lying to the left of the section being considered ; i.e. 

A'C = C'Odx. 
Substituting in Eq. 1, 

dy 



SA'- Codx 



dx D' 

Differentiating and dividing by dx, we have, neglecting signs, 

(Py_<l 

dx 2 ~I)' 

The general equation of the elastic curve (Lanza, p. 301) is 



(2) 



dx*~EI ' ■' ' ' •' " ' ' K) 

In order for the curve (Fig. 42 B) to be the elastic curve of the 
beam, the second members of Eqs. 2 and 3 must be equal. 

...0 = M = £D DD , = EI . . . . (4) 

D' EI MI w 



DEFLECTION OF BEAMS. 59 

Let a = ratio of true to plotted dimensions of the beam. Then, 
in order for the ordinates of Fig. 42 B to represent the true deflec- 
tion of the beam, the vertical scale of Fig. 42 B must be increased a 
times. For this purpose D' must be reduced in the same ratio 
(Art. 20) ; also, if we wish to exaggerate the deflections in any 
ratio n, D' must be reduced proportionally. In this latter case 
Eq. 4 becomes 

DD' = — (5) 

By means of Eq. 5 suitable values of D and D f can be deter- 
mined such that the ordinates of the second funicular polygon 
(Fig. 42 B) shall represent the deflection of the beam exaggerated 
n times. 

It is not necessary for SS' (Fig. 42 B) to be horizontal, since the 
ordinates are constant so long as the pole distance D' is not altered. 

If the bending moment changes sign as in Fig. 39, the areas cor- 
responding to negative moments must be plotted in Fig. 42 B f in the 
opposite direction from the positive ones. 

If the beam is of non-uniform section, Eq. 5 shows that DD' 
must vary in the same ratio as I. The proper elastic curve for this 
case is then constructed by making the pole distance D' vary in the 
same ratio as / varies. 

To determine the deflection at any given section of the beam, 
make the section one division line of the area (Fig. 42 A). In other 
respects the area may be subdivided in any manner (see Art. 68), 
with the exception that when the beam is of non-uniform section 
the area must be divided at those sections where the value of 1 
changes. 

71. Example 1. Fig. 1, Plate III, is a cantilever, 10 ft. span, 
loaded at two points, i" has three different values, as given in the 
diagram. E = 30000000 lbs. per sq. in. It is required to draw 
the moment diagram and the elastic curve. 

Scale of space diagram, 1 : 20. .-. a = 20. Scale of force dia- 
gram (Fig. la'), 4000 lbs. = 1". The pole distance PA is taken to 
be 4". .-. D = 4 x 4000 = 16000. The construction of the moment 
diagram (Fig. la) needs no explanation. Its surface is divided as 
shown and the centres of gravity are indicated by circles. The 
areas of these divisions are plotted in Fig. lb' to the scale, 
1000 sq. in. (full size) = 1". The full size area is obtained by multi- 
plying the diagram area by a 2 =(20) 2 = 400. The deflections are to 
be exaggerated 5 times, i.e. n = 5. 



60 NOTES ON GRAPHIC STATICS. 

rrr 

Substituting the preceding values in the formula DD' — — , we 

have 16000 D' = 3QQ00Q °Q ' 2Q0 . ... D' = 3750 = 3.75" to the scale 

5 • 20 

(1000 sq. in. = 1"). The pole P' is taken with a pole distance of 
3.75", and the elastic curve (Fig. lb) is constructed. The strings 
corresponding to P'G', P'F', P'E' are drawn in order. At the sec- 
tion e' the value of I changes to 150, and the pole distance must be 
reduced in the same ratio, the new pole P" lying on P'E'. The 
construction is continued in a similar manner, the pole for the 
portion of the beam where 7=100 being P'". The ordinate mn, 
divided by 5, determines the deflection of the beam to be .496". 
The computed deflection is .493". 

Example 2. The beam (Fig. 2, Plate III) is supported at two 
points, a and h, 14 ft. apart, and overhangs 6 ft. at the right end. 
It is loaded with a distributed load of 300 lbs. per foot. Cross 
section = 6" x 12". /= 864. E = 1200000. Scale, 1 : 40. 

The load is divided into 10 equal lengths, each division being 
concentrated at its middle point. Scale of force diagram (Fig. 2 a'), 
2000 lbs. = 1". Pole distance = 1£" ; i.e. D = 3000. The surface 
of the moment diagram (Fig. 2 a) is divided as shown and the areas 
plotted in Fig. 2 b' to the scale, 500 sq. in. = 1", the positive and 
negative areas being laid off in opposite directions, n = 5. 

EI 

Substituting in the formula DD' = — , we have 

na 

3000 ^ = 1200000. 864 = 1728 . 
5-40 

i.e. the pole distance (Fig. 2 b') is -\\^ = S"A56. Fig. 2 6 is the 
elastic curve. The measured ordinates at 7 ft. and 20 ft. from the 
left end divided by 5, give for the deflections at these points — .144" 
and + .012". The computed deflections are — .14" and -f .01". 



§ 3. Beams supported at More than Two Points. 

72. Determination of Reactions. In determining the reactions 
of the supports for such cases, the theory of elasticity must be 
employed, no solution being possible by means of the methods of 
statics of rigid bodies. From the theory of beams we know that the 
deflection at any point due to any system of loads is equal to the 
algebraic sum of the deflections produced by the loads taken sepa- 
rately. A method of determining the supporting forces, based on 
this fact, is illustrated by the following example : 



BEAMS SUPPORTED AT MORE THAN TWO POINTS. 



61 



The beam (Fig. 43) of uniform section is supported at four equi- 
distant points, all on the same level, as shown, and loaded uni- 
formly, the total load being W. It is required to find the reactions 
of the supports. 



. n L b iM d IM f I.?l h l i l J ..] l !l l l w ! n i t" 




Fig. 43. 



If the intermediate supports R and R' were removed, the beam 
would deflect as shown by the dotted curve. The deflections at the 

supports would then be v =v' = (Lanza, p. 303, Eq. 2). 

y7Z Mil 

The reactions R and R' must evidently be equal to the forces which 
would raise these points to their original positions, i.e. would pro- 
duce upward deflections at R and R' equal to v and v' respectively. 
The deflections at the intermediate supports due to the force 

R are "* = dS and v '° = iiSr (Lanza ' p - 308 ' Eq - 2> The 

deflections at these points due to R' are 



v„- = ^rr r and v' R > = 



486 EI 



243 EI 



62 NOTES ON GRAPHIC STATICS. 

-rr ±Rl s , 7iZ7 3 11IF7 3 

Hence, v=v R +v R ' = —— = 

B * 243EI AS6EI 972EI 

Similarly, v' = v' B -\- v'^ = — 1- 



486^/7 243 #7 972 777 

Solving for R and 72', we obtain R = R' = ^ W. The two 
remaining supporting forces can now be found by moments. In 
this case each equals -^ W. The general method of determining 
supporting forces, illustrated by the preceding example, is of wide 
application, but in any complicated case is laborious. 

73. Construction of Moment Diagram. The load W is divided 
into 15 equal divisions. These loads and the reactions previously 
found are plotted in order in the force diagram, and with P as pole, 
the moment diagram (Fig. 43 B) is constructed. The moments are 
negative in the spaces t-t' and t"-t'", these four points being the 
points of inflexion of the elastic curve. 

This diagram can better be constructed as follows : So long as 
the pole distance remains constant, the intercepts of Fig. 43 B will 
not be altered; hence different poles can be used in constructing 
different parts of the funicular polygon, provided the pole distances 
are all equal. In Fig. 43 A the moment diagram is constructed, 
using the poles, P', P", P'", for the strings lying in the three spans 
respectively. The pole distances are reduced one-half, hence the 
intercepts will be double those of Fig. 43 B (Art. 20). This use of 
different poles gives a much more satisfactory diagram than when 
a single pole is used. 

74. Second Method of constructing Moment Diagram and Deter- 
mining Reactions. Eef erring to Fig. 43 A, the funicular polygon 
xmm'x' for the loads can be drawn without knowing the supporting 
forces. In order to complete the moment diagram it is necessary to 
know the values of the intercepts mn and m'n'. These intercepts 
are equal to the bending moments at these points divided by the 
pole distance (Art. 18). Hence the points n, n' can be located and the 
diagram completed when the bending moments at the intermediate 
supports are kDOwn. These moments, can be computed by means of 
the " three-moment equation." (For the method of determining the 
bending moments at the supports of a continuous girder by means of 
the " three-moment equation," see Lanza's Applied Mechanics, Chap. 
VIII.) Having completed the moment diagram by drawing the 
strings xn, nn ( , and n'x 1 , the reactions UQ, QB, RS, SA, are deter- 



BEAMS SUPPORTED AT MORE THAN TWO POINTS. 



63 



mined by drawing the rays P'S, P"E, P'"Q, parallel to these 
strings. 

From the moment diagram, the bending moment at any point can 
be determined, and the elastic curve constructed in the manner pre- 
viously described. 

The values of the reactions of the supports (or bending moments 
at the supports) for the case of a continuous beam of uniform sec- 
tion, equal spans, and loaded uniformly over its entire length, will 
be found in text-books on the subject. The stresses in continuous 
beams are quite uncertain, as they are altered by unequal settling of 
the supports. 

75. Centre of Gravity. The following constructions for centre 
of gravity will be found useful in connection with the work of this 
and the following chapter. 

1. Centre of Gravity of Any Quadrilateral. 
Two constructions are given in Fig. 44: 
(1) Bisect each diagonal. Connect the point 
of intersection of the diagonals with the mid- 
dle of the line DE, joining their middle points. 
The centre of gravity will be at the extremity 
of this line when extended one-third of its 
length. (2) Lay off BC = OA and join C with the middle, D, of the 
other diagonal. The centre of gravity lies on this line one-third of 
CD from D. X 



2. Centre of Gravity of a Trapezoid. Two 
constructions are given in Fig. 45 : (1) Draw 
the medial line GH and one diagonal. Bisect 
the medial line at K. The distance from the 
point of intersection L to K, prolonged one- 
third of its length, will locate the centre of 
gravity. (2) Lay off AF= CD, and DE=AB. 
The intersection of EF and the medial line is 
the centre of gravity. 




Fig. 44. 




Fig. 45. 



CHAPTER IV. 

MASONRY ARCHES, ABUTMENTS, ETC. 

§ 1. General Conditions of Stability. 

76. Nature of the Forces involved. Let PRMN (Fig. 46) be a 
block of masonry acted upon by a force AB. In addition to this 
force, the weight of the block must be taken into account. This 
weight is represented by BC, its line of action be being drawn 
through the centre of gravity of the block. The resultant of these 
two forces is AC, its line of action passing through the intersection 
of ab and be. AC is, therefore, the resultant pressure exerted by 
the block upon the plane MN. This plane may be taken to be a 




M 




N 
Fig. 46. 




Fig. 47. 



joint of the masonry or its base. Also, the forces which hold the 
block in equilibrium are AB, BC, and the reaction of the plane MN, 
this last being a force equal and opposite to AC. Moreover, AC 
represents the resultant stress on the plane MN. 

Again, let PRMN (Fig. 47) be an arch stone, BC being the load 
supported by this stone including its own weight. The line of 
action be of this load passes through the centre of gravity. BC is 
balanced by the forces AB and CA exerted upon PRMN by the 
adjacent arch stones. These three forces must, therefore, form 
a triangle, and their lines of action must intersect at the same 
point. 

64 



GENERAL CONDITIONS OF STABILITY. 65 

77. Resistance of a Masonry Joint. The conditions of stability 
for the block of Fig. 46 as far as the joint MN is concerned are evi- 
dently the following : (1) the block must not overturn about an edge, 
as N\ (2) it must not slide over the joint ; (3) the material of the 
stone and mortar must not crush. These three conditions will be 
discussed in turn. 

78. Resistance to Overturning. In this connection, the tensile 
strength of the mortar joint is commonly neglected. Then the 
block (Fig. 46) would evidently overturn, if the line of action of 
the resultant force AC pierced the plane MN outside of the surface 
of the joint. The moment of AC about N as moment axis is the 
measure of the resistance to overturning about this edge, i.e. in order 
to overturn the block it would be necessary to apply a force whose 
moment about N was equal to that of AC, but having the opposite 
sign. 

79. Resistance to Sliding. Let the resultant force AC (Fig. 46) 
be resolved into components parallel and perpendicular to the joint, 
as indicated. The normal component represents the direct pressure 
on the joint, while the parallel component tends to slide the block 
over the joint, and must be resisted by the sliding friction at the 
joint, the adhesion between the stone and mortar being neglected. 

Coefficient of Friction. Let P (Fig. 48) be the resultant pressure 
of the block on the plane AB, and <£ the minimum angle of inclina- 
tion with the normal at which sliding will occur. This angle <£ 
is called the angle of repose, and tan <£, or ratio of tan- 
gential to normal component of the force, is called the 
coefficient of friction /. It is shown by experiment 
that / is practically constant for given surfaces, i.e. is 
independent of the intensity of the normal pressure. 
In the case of masonry joints, the minimum value of Fm ' 48- 
the coefficient of friction is taken to be from .4 to .5. In order, 
then, for sliding not to occur, the resultant pressure at any joint 
must make with the normal an angle less than tan -1 .4. 

80. Resistance to Crushing. The normal component of AC 
(Fig. 46) represents the resultant compression stress at the joint 
MN. This stress is assumed to be uniformly varying. (See Lanza, 
p. 265.) The three cases which may occur are represented in Fig. 
49. In Fig. 49 A the stress is distributed over the whole surface 
of the joint, the limiting case being Fig. 49 B, where the neutral axis 




m 



NOTES ON GRAPHIC STATICS. 



is at one edge M. 
(A) 

[f 



M 



(B) - 

^tTlllTllllll. . 



M 



(C) 



,i\ 



In Fig. 49 C the pressure is distributed over the 
portion XN of the joint. If the joint were capa- 
ble of resisting tension, this last would be the 
case where the stress is partly tension and partly 
compression, but, assuming the joint -incapable 
of resisting tension, the portion MX is without 
stress and tends to open. 

If the surface of the joint is rectangular, the 
resultant stress R, in Fig. 49 B, acts at -| MN 



M 



Fig. 49. 



from M. 
Let 
I 



In the formula x l = 



x A 



Proof: 
MN= h = depth of joint. 

(Lanza, p. 265) we have 



I=\bh? = ±Ah 2 , x Q = \h. 

^Ah 2 
Substituting, x ± = ^ = -|A. 
2- An, 

In Fig. 49 A the resultant stress lies nearer the middle of the 
joint than in Fig. 49 B; hence, In order for the pressure to be distrib- 
uted over the whole surface of a rectangular joint, the resultant must 
act within the middle third of its depth. The corresponding limits 
within which the resultant pressure must act, in case of any other 
form of surface, can be found in a similar manner. 

When the stress is distributed over the whole surface of the 
joint, its maximum intensity can be found by the formula for short 
struts (Lanza, § 207). When the stress extends over a portion of 
the joint, as in Fig. 49 C, the short strut formula is evidently inappli- 
cable. In such a case, when the surface is rectangular, the point X 
can be found by making NX = 3NR, R being the point of applica- 
tion of the resultant stress. Its maximum intensity is then equal to 

— ■. If R acts at the edge N, the area NX becomes zero and 

area NX 

the intensity of the stress infinity. 

For safety the maximum stress must not exceed the working 

compression strength of the material. 



§ 2. Masonry Arch. Line of Pressure. 

81. Definitions. In Fig. 50, a and b are respectively the span 
and rise of the arch, h is the thickness of the arch ring, also the depth 
of the joints of the arch ring. The highest part of the arch is the 



MASONRY ARCH. LINE OF PRESSURE. 



67 



crown. The portions of the arch ring between the crown and abut- 
ments are the haunches. The arch stones are also called voussoirs. 
The inner surface, cmc, of the arch ring is the intrados. The outer 
surface, dm'd, is the extrados. These names are also given to the 





A 


-ssc 


m> 


> 




U£? 


< 


y 


^ 


. i_ 


"X 


y 


c 




-a 


c 













Fig. 50. 



corresponding curves. The lines c, c, where the intrados meets the 
abutments, are the springing lines. 

Arches are designated according to the form of the intrados. A 
full centred arch is one whose intrados is a semicircle. A segmental 
arch is one whose intrados is a circular arc of less than 180°. A 
pointed arch is one where the two half-curves of the intrados inter- 
sect at the crown instead of forming a continuous curve. The 
wooden frame which supports the arch during construction is the 
centre. (For further definitions see Baker's " Masonry Construc- 
tion.") 

82. Line of Pressure, a Funicular Polygon. Let AB, BC, CD, 

etc. (Fig. 51), be the loads supported by the arch stones. If the 
pressure at any joint, as a, is given 
completely, the pressures at the other 
joints can be found by the triangle of 
forces (Art. 76). Thus, representing 
the pressure on the joint a by PA, the 
resultant of PA and AB, i.e. PB, will 
be the pressure on the joint b, its line 
of action passing through the inter- 
section of pa and ab, as shown. Simi- 
larly, PC is the pressure on the joint c, 
PD on the joint d, etc. Thus it is 
seen that the lines of action of the 
resultant pressures on the succes- 
sive joints of an arch are the strings of a funicular polygon, the 
corresponding rays representing the magnitudes of these pressures. 




Fig. 51. 



68 



NOTES ON GRAPHIC STATICS. 



This funicular polygon will be referred to as the line of pressure of 
the arch, although the line of pressure or line of resistayice, as com- 
monly defined, is the broken line joining the centres of pressure of the 
successive joints. 




Fig. 52. 



83. Symmetrical Arch. Symmetrically loaded. In this case it is 
evident that the line of pressure will also be symmetrical with refer- 
ence to a vertical through the crown, and hence the pressure at the 
crown will be horizontal. Only one-half the arch ring need then be 
considered. 

Let PA (Fig. 52) be the 
horizontal crown pressure, AB, 
BC, etc., being the loads. The 
funicular polygon constructed 
by using P as pole is then the 
line of pressure for the half- 
arch. 

84. Test of Stability. A 

common test for the stability 
of a proposed arch, as far as 
its outline is concerned, is to 
determine the possibility of 
drawing a line of pressure which will lie wholly within the middle 
third of the thickness of the arch ring. If the resultant pressure 
at each joint acts within the middle third of its depth, the com- 
pression stress will be distributed over the whole surface of the 
joint (Art. 80). The general method of determining the possibility 
of drawing a line of pressure within the middle third of the arch 
ring is illustrated in Fig. 53. 

The arch is full centred. The two curves drawn include 
between them the middle third of the thickness of the arch ring. 
The loads AB, BC, etc., are taken to be vertical. Assume any pole 
P on a horizontal line through A and construct the funicular polygon 
mn. Next assume two points, as X and Y, in the arch ring, and 
construct a funicular polygon to pass through them. To do this we 
know that the reactions at any two joints, as X and Y, must balance 
the resultant load included between these joints. This resultant 
load It acts through the intersection of the strings a and g of the 
polygon mn. The corresponding strings of the desired polygon must 
then intersect on R. The string a, passing through X, is horizontal ; 
the string g, therefore, has the direction YO. The pole P' of this 



MASONRY ARCH. LINE OF PRESSURE. 



69 



polygon is now located by drawing the ray GP' parallel to YO, and 
the polygon is constructed. This polygon XY departs farthest from 
the middle third of the arch ring at the joint e. A polygon whose 
strings a and e pass through X and Y' respectively is then most 




likely to fall within the required limits. The pole P" of this poly- 
gon is located in a manner similar to P'. This last polygon XYY" 
lies within the designated limits except near the springing plane, 
where the requirement is not essential in the case of full centred 
arches, since this portion of such an arch can be treated as part of 
the abutment. 

85. Maximum and Minimum Crown Pressure. It is evident that 
as the pole distance (i.e. pressure at the crown) (Fig. 53) increases, 
the rays and corresponding strings become more nearly horizontal, 
and consequently the funicular polygon becomes more nearly flat. 
Hence of all lines of pressure which can be drawn within the middle 
third of the arch ring, that one corresponds to a minimum crown 
pressure which touches the outer curve at or near the crown and the 
inner curve at some point farther from the crown, while the line of 
pressure corresponding to a maximum crown pressure will touch 
the inner curve at or near the crown and the outer curve at some 
point farther from the crown. 

86. Position of True Line of Pressure. An infinite number of 
funicular polygons may be constructed for a given system of arch 



70 NOTES ON GRAPHIC STATICS. 

loads. In general it is uncertain which of these is the true line of 
pressure. Of the various theories (see references) which have been 
advanced to serve as a basis for determining the position of the true 
line of pressure, the theory of least crown thrust seems to be most 
commonly employed. It is, therefore, briefly presented here. 

This theory is essentially that the true line of pressure is that 
one which, lying within the middle third of the arch ring, cor- 
responds to a minimum crown pressure. It appears to be based 
upon the observation that most arches settle at the crown when the 
centre is removed, and upon the assumption that the crown pressure 
is a passive force developed by the tendency of the two half-rings to 
tip towards each other, and is the least force that is necessary to 
prevent such overturning. 

If the arch settles at the crown (as a result of rotation, not slid- 
ing, of the arch stones), the obvious tendency will be to open the 
joints as shown in Pig. 54, the resultant pressure moving upward at 





Fig. 54. Fig. 55. 

the crown and inward at the haunches. This position of the line of 
pressure corresponds to minimum crown thrust (Art. 85). The 
joints a, a, where the tendency to open at the extrados is greatest, 
are called the joints of rupture. They correspond to the points where 
the line of pressure touches the inner limiting curve (Fig. 53). In 
the case of a full centred arch or elliptical arch, the joints of rupture 
are approximately 30° from the horizontal. In a segmental arch 
subtending less than 120°, they are at or near the springing planes. 
In the case of a pointed arch, or an arch very lightly loaded at 
the crown and heavily loaded at the haunches, the tendency may be 
for the crown to rise and the haunches to move inward (Fig. 55). 

87. Construction of True Line of Pressure according to Theory of 
Least Crown Thrust. In Fig. 53 XY'Y" is the true line of pressure 
according to this theory, Y' being at the joint of rupture. The 



MASONRY ARCH. LINE OF PRESSURE. 



Tl 



trial line of pressure may be drawn at once through X and Y', the 
joint of rupture being located as explained in Art. 86. If the line 
of thrust thus drawn does not fall within the middle third, the con- 
struction can be repeated as explained in Art. 84. 



88. Example. Figure 1, Plate IV, is one-half of a symmetrical 
full centred arch in a masonry wall whose height is limited by a 
horizontal line, as shown. It is required to draw the line of press- 
ure according to the theory of least crown thrust, and to determine 
if the arch satisfies the conditions of stability. 

The half-ring is divided by radial lines, which need not coincide 
with the actual joints of the arch. It will be assumed that each of 
these divisions or voussoirs supports the weight of the portion of 
wall directly above it, as indicated by the vertical lines. If the 
specific gravity of the material above the arch ring is the same as 
that of the arch ring, the load supported by any voussoir, as mn, is 
proportional to the area of the polygon mnn'm'. If these specific 
gravities are unequal, the vertical orainates may be altered in 
length so that the areas above the voussoirs will represent weights 
to the same scale as the areas of the voussoirs themselves. Other- 
wise the weights of the voussoirs and material above them may be 
dealt with separately. In this example, the wall is of uniform thick- 
ness and the weight of the masonry is, throughout, 160 lbs. per cubic 
foot. 

Considering one foot thickness of wall, the loads (see table) are 
calculated by multiplying the corre- 
sponding areas by 160 lbs. HI is taken 
to include two voussoirs to avoid con- 
fusing the drawing ; IJ is the* weight of 
the masonry to the right of the line i ; 
JK is the weight of the masonry be- 
low RS. 

The resultant loads act at the centre 
of gravity of the areas. The centre of 
gravity of the voussoir mn is 0\ and the 
centre of gravity of the trapezoid above 
this voussoir is 0" (see Art. 75). The 
centre of gravity of the entire area 
mnn'm' is then found by dividing the 
line O'O" into parts inversely propor- 
tional to these areas. The centres of gravity are indicated by 
circles. 



Table of Loads. 
(Fig. 1, Plate IV.) 



Voussoir. 


Wt. of 
Voussoir. 


Wt. above 
Voussoir. 




lbs. 


lbs. 


AB 


540 


1360 


BC 


540 


1440 


CD 


540 


1570 


BE 


540 


1700 


EF 


540 


1790 


FG 


540 


1760 


GH 


540 


1540 


HI 


1080 


1600 


IJ 


— 


10240 


JK 


— 


7680 



72 NOTES ON GRAPHIC STATICS. 

The line of pressure is now constructed as follows : The loads 
AB, BC, etc., are plotted to scale, and, selecting any pole P on a hori- 
zontal line through A, a trial polygon xy is drawn, the point x being 
one-third the depth of the joint below the extrados. In drawing this 
polygon, the intersections c', d', etc., of its various strings with the 
string a are marked. These points of intersection locate the result- 
ant load lying between the two intersecting strings; hence, the 
strings of any other polygon for the given loads will pass respectively 
through these same points, if the first string a is unchanged. The 
line of pressure desired is such that the resultant pressure at the 
joint of rupture will act at one-third the depth of the joint from the 
intrados. The joint of rupture may be determined by trial, as 
follows : Knowing that it is about 60° from the crown, we trisect the 
joint g at 1, and draw the string g through 1 and g'. The adjacent 
strings are then drawn, and it is thus found that / is the true joint 
of rupture. This joint is then trisected at 2, and the string 2/' is 
drawn. The polygon is completed by drawing the remaining strings 
in succession through e', d', etc. The pole P' of this polygon is 
located by drawing the ray FP* parallel to the string /. The 
points of application of the resultant pressure at the different 
joints are indicated by arrows. This line of pressure falls outside 
the middle third of the joint, at the springing plane and first joint 
above. This portion of the arch can be treated as part of the 
abutment. 

Aside from the condition that the pressure must act within the 
middle third of the arch ring, the resistance to sliding and crushing 
must be investigated. 

As regards sliding, it is seen that the direction of the resultant 
pressure at each joint is very nearly normal to the joint, with the 
exception of the springing plane, where the pressure p'i makes 
with the normal an angle greater than tan -1 .4. When, however, the 
weight ij is combined with pi, the resultant pressure, p"j, satisfies 
the requirement for safety against sliding. 

For crushing, the maximum compression stress is to be calculated 
at each dangerous joint. For example, the resultant stress on the 
joint /, found by scaling off the ray P'F, is 13100 lbs. The area 
of the surface of the joint is 1 x If = If sq. ft. = 252 sq. in. Aver- 
age pressure per sq. in. = ^fll-- = 52 lbs. Hence the maximum 
stress = 2 x 52 = 104 lbs. per sq. in. (See Art. 80.) This stress 
must not exceed the working compression strength of the masonry. 
(For strength of masonry, see Lanza's Applied Mechanics and other 
references.) 



MASONRY ARCH. LINE OF PRESSURE. 73 

89. Unsymmetrical Cases. When the arch or loading is unsy in- 
metrical, the line of pressure is also unsymmetrical, and must there- 
fore be drawn for the whole arch. The construction of the line of 
pressure involves the problem of drawing a funicular polygon through 
three points. (See Arts. 23 and 24.) 

Example. (Fig. 2, Plate IV.) Given a segmental arch of 16 ft. 
span and 3 ft. rise. Thickness of arch ring = 11 ft. The left and 
right halves are loaded with 3200 lbs. and 6400 lbs. respectively, 
these loads being uniformly distributed over the arch ring. It is 
required to determine the possibility of drawing a line of pressure 
within the middle third of the arch ring. 

The arch is divided into 16 equal voussoirs, and the load sup- 
ported by each is assumed to act at the middle of its outer surface. 
The loads are plotted to scale, and the funicular polygon xy is con- 
structed, using P for pole. Selecting the points V, v', and u' at 
one-third and two-thirds the depth of the joints from the intrados, 
the funicular polygon which will pass through these points is 
located. This polygon is not drawn, but the points on it, falling 
outside the middle third, which serve to locate the final line of press- 
ure, are marked by circles. These points were determined by the 
method of Art. 24. It is seen that, to the right of the crown, this 
polygon rises above the middle third, while near the left abutment 
it falls below. From the position of these points it appears that a 
polygon drawn through the three points V, 2', and v' will probably 
fall within the specified limits. The pole P' of this final polygon 
was located by the method of Art. 23. PZ and PZ' are drawn 
parallel to 1, 2 .and 2, v respectively ; then ZP' and ZP' are drawn 
parallel to 1', 2' and 2', v' respectively. The point of intersection 
of these lines is P'. From P' as pole the polygon is constructed so 
that the strings e, k, and s pass respectively through the points 1', 
2', and v'. This polygon lies wholly within the middle third of the 
arch ring. 

The location of this polygon corresponds to that of the true line 
of pressure of a symmetrical arch according to the theory of least 
crown thrust, with the exception that it does not touch the outer 
limiting curve at the crown. Safety as regards sliding and crushing 
is investigated as previously explained. 

90. General Remarks. (See also Lanza's Applied Mechanics, 
§ 270.) The investigation of the stability of a voussoir arch is 
necessarily inexact. 

(1) The loads are more or less uncertain, both in amount and 



74 



NOTES ON GRAPHIC STATICS. 



distribution. In the example (Fig. 1, Plate IV) it was assumed that 
each voussoir supported the weight of the entire mass above it. 
This is a common assumption in dealing with bridge arches. In 
some cases such an assumption would be manifestly absurd ; e.g. an 
arch in a high masonry wall, or a tunnel arch deep under ground. 

(2) The location of the true line of pressure is uncertain. The 
method already explained for locating the line of pressure is con- 
ventional. It is not derived mathematically. It has been proposed 
by different writers to place the determination of the line of press- 
ure upon a rational basis, by treating the masonry arch as an 
elastic arch fixed at the ends. It is not fully evident that the 
voussoir arch conforms sufficiently closely to the case of an elastic 
arch with fixed ends to make such treatment reliable, even assuming 
that the loads are accurately known. (For discussions of arch 
theories, see References.) 

The stability of the abutments is essential to that of the arch, 
and must be considered in connection with the arch. 



§ 3. Abutments, Piers, etc. 



91. Conditions of Stability. The general conditions of stability 
of § 1 are applicable to any piles of masonry subjected to the action 
of external forces, such as the thrust of an arch or truss, pressure of 
earth, water, wind, etc. 



92. 

zontal 
acting 



Example 1. Fig. 56 is an abutment subjected to a hori- 
pressure AB and a vertical pressure BC, their resultant, AG, 
at the point C. CD, DE, EF, and FG are the weights of the 

four divisions of the masonry, their 
centres of gravity being marked by 
circles. The pressure on the joint 
d is the resultant of AC and the 
weight CD of the first block. This 
resultant is AD, its line of action 
passing through the intersection of 
ac and a vertical through the centre 
of gravity of the block. The point 
of application of this resultant press- 
ure, at d, is indicated by an arrow. 
The resultant pressures at the re- 
maining joints are found in a simi- 
Fig. 56. lar manner. 





a \ 


b 


c 








\ 


b X 


C 




a 


\ 




\ 


B 




\ 


-c 


K 


d 




\ 


\ 


\ 




\ 




o 


K\ 




\ 


-D 
-E 




V 








6 


\ 






-F 




\\f 














6 \ 




-G - 


\,Q 



ABUTMENTS, PIERS, ETC. 



75 



The conditions of safety as regards sliding, overturning, and 
crushing have been previously discussed. The maximum pressure on 
the soil must also be kept within safe limits. (See Baker's Masonry 
Construction, Chap. X.) 

A broken line connecting the points of application c, d, e, f and 
g of the resultant pressures on the successive joints of an abutment 
is called the line of resistance or line of pressure, as in case of an 
arch. (See Art. 82.) 

93. Example 2. The abutments of a masonry arch can be 
considered in connection with the arch. In Fig. 1, Plate IV, the 
weights IJ and JK, lying above and below the joint US, are plotted 
to half scale (4000 lbs. = 1 in.), the pole for these two loads being 
located by bisecting the ray P'l at P". The resultant pressure on 
BS is P"J, its line of action p"j passing through the intersection 
of p'i and if The resultant pressure on the base is P"K, its line of 
action being p"k. To find the maximum intensity of the pressure 
on the base we have P"K= 9.1 (inches) x 4000 = 36400 lbs. By 
measurement, p"k acts 1.1 ft. from the centre of the base. The 
bearing area is 7f- x 1 = 7| sq. ft. Substituting these values in the 



formula p = 



P , Px n a 



(Lanza, § 207), the maximum intensity of the 



A I 

pressure is found to be 60 lbs. per sq. in. 

94. Example 3. Let Fig. 57 represent a pier supporting the 
thrust of an arch on each side. These 
thrusts are AB and BC, their point 
of intersection O lying on the centre 
line of the pier. Let CD, DE, EF, 
and FG represent the weights of the 
pier divisions, CD being the weight 
of the pier masonry above the joint 
d. The resultant pressure on d is 
AD, its line of action Od passing 
through O. The lines of action of 
the pressures on the remaining joints 
will also pass through O, since the 
vertical through this point contains 
the centres of gravity of all the 
blocks. These lines, are Od, Oe, Of, 
and Og, drawn parallel respectively 
to AD, AE, AF, and AG. If 
the thrusts of the two arches are equal and equally inclined, the 




-E 



7'- LA 
t 




I if 


1 ig 



Fig. 57. 



76 



NOTES ON GRAPHIC STATICS. 



resultant pressure on the pier will evidently be vertical and equal 
to the weight of the two half -arches besides the weight of the pier 
itself. 

95. Example 4. Let Fig. 58 represent a chimney subjected to 
wind pressure. The weights of the portions ab, be, etc., are AB, 
BC, etc., and the wind pressures on these portions are AB', B'C, 
etc. The lines of action of these wind pressures are the horizontal 



F'E'D'C'B' 

l 1 I l l ,A 





lines ab, be, etc. With any pole P draw the funicular polygon mn. 
(For this purpose the wind pressures should be plotted to a larger 
scale.) 

To find the resultant pressure at any section, as e, the line of 
action tr of the resultant wind pressure above that section is located 
by the intersection of the strings a and e. 

The pressure at the section e is the resultant of this wind press- 
ure and the weight of masonry above e. The line of action rs 
of this pressure will act through r, the point of intersection of the 
resultant wind pressure and weight of masonry above e, and its 
direction will be parallel to E'E, its magnitude being represented by 



ABUTMENTS, PIERS, ETC. 77 

the length of E'E. The resultant pressure at any other section can 
be determined in a similar manner. 

The preceding examples will serve to indicate the method of 
determining the stability of such structures when the loads are 
known. 



REFERENCES. 



The following list includes elementary works in which more detailed 
explanations will be found, and others of a more advanced character. The 
list is not intended to be exhaustive. 

" Graphic Statics," by Mansfield Merriman. John Wiley & Sons, New 

York. 
" Elements of Graphic Statics," by L. M. Hoskins. Macmillan & Co., New 

York. 
"Construction of Trussed Roofs," by N. C. Ricker. W. T. Comstock, New 

York. 
"Statique Graphique," by Rouche. Baudry & Co., Paris. 
" Charpentes Metalliques," by Dechamps. Yaillant-Carmanne, Liege. 
"Theory and Practice of Modern Framed Structures," by J. B. Johnson. 

John Wiley & Sons, New York. 
"Applications de la Statique Graphique," by Koechlin. Baudry & Co., 

Paris. 
" Treatise on Masonry Construction," by I. O. Baker. John Wiley & 

Sons, New York. 

For additional illustrations of roof construction, see 

" Revue Technique de L'Exposition de Chicago," Part I, Architecture (with 

Atlas). E. Bernard & Co., Paris. 
Publications of Iron Construction Companies. 



Ficj.2.(See p. ). 



PLATE I. 
Fig. I (See p. ) . 




Fig. 2b. Wind Roller Sid 



Scales. Fig 
5ft.= I inc 



' ii i i r 



5 

3ooolbs.= ih 



i. i i 



3ooo 





Fiy-Zd^ Stresses. 




Fig. 3 b 

Wind Right. 




Fig. 3 (See p. ) 

Scales. 
/oft.= I inch. 



Plate nr. 



Fig. 2 

d i b i c i d i e i f l ? i S l I J i 





Fig. 3b 
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Fiq.l (Seep. 
Scales. 
2ft.= I inch, 



2 

20 00 lbs = fine 

i : 

2000 



PLATE N. 



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*B 




Fig. I (See p. ) 

Scales. 

2ft.= I inch. 



2ooo!bs.=linch 



2000 4000 6000 




Fig. 2 (Seep. ) 

Scales. 

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